SECOND  YEAR 
COLLEGE  CHEMISTRY 

A  Manual  of  Laboratory  Exercises 


BY 

WILLIAM  H.  CHAPIN 

ASSOCIATE    PROFESSOR    OF    CHEMISTRY    IN    OBERLIN    COLLEGE 


NEW  YORK 

JOHN  WILEY  &  SONS,  Inc. 

LONDON:  CHAPMAN  &  HALL,  LIMITED 
1922 


A 


4  g 


Copyright,  19122 

BY 

WILLIAM  H.  CHAPIN 


PRESS  OF 

BRAUNWORTH   &  CO. 

BOOK   MANUFACTURERS 

BROOKLYN,    N.    Y. 


PREFACE 


THE  work  outlined  in  this  manual  is  intended  to  accompany  the 
author's  text,  "  Second  Year  College  Chemistry."  It  may,  according 
to  the  author's  experience,  be  used  in  connection  with  a  course  in  quan- 
titative analysis.  The  general  order  of  procedure  in  both  classroom 
and  laboratory  has  been  outlined  in  the  preface  to  the  text. 

All  the  experiments  here  presented  have  been  carefully  tested  out 
with  several  classes  of  students  and  then  carefully  rewritten.  The 
author  feels,  therefore,  that  careful  following  of  directions  will  insure 
good  results.  In  all  work  of  a  quantitative  nature,  however,  neatness 
and  natural  aptitude  are  prime  factors  which  cannot  be  forced  upon  the 
student.  The  student  who  is  naturally  sloppy  and  clumsy  will  prob- 
ably emerge  from  any  course  with  his  original  traits,  and  will  never 
measure  up  as  a  real  chemist.  However,  neatness  and  skill  on  the  part 
of  the  teacher  are  also  necessary,  for  even  a  good  student  may  be  ruined 
by  lax  supervision  and  a  careless  example. 

The  apparatus  employed  in  the  experiments  is  made  as  simple  as 
possible.  This  has  been  done  for  two  reasons:  first,  the  author 
desires  to  make  it  possible  to  work  through  this  entire  manual  in  places 
where  the  cost  of  more  elaborate  apparatus  would  be  prohibitive;  and 
second,  he  desires  to  emphasize  principles  rather  than  the  bewildering 
details  of  refined  methods  and  instruments.  On  the  other  hand,  the 
aim  has  been  constantly  kept  in  mind  to  make  the  work  accurate  and 
truly  quantitative.  Careful  directions  have  been  included  for  the 
preparation  of  all  the  apparatus,  although  this  preparatory  work  need 
not  always,  be  done  by  the  student  where  the  time  is  limited. 

One  of  the  most  important  parts  of  successful  laboratory  work  is  the 
careful  keeping  of  a  notebook.  The  teacher  is,  therefore,  urged  to 
insist  that  neat,  accurate  notes  accompany  each  experiment.  These 
notes  should  be  taken  in  the  permanent  notebook  and  in  their  final  form, 
as  soon  as  the  data  are  obtained.  The  habit  of  writing  notes  on  scraps 
of  paper  and  afterwards  copying  them  off  in  "  better  form  "  is  pernicious. 
Proper  entry  of  notes  should  be  regarded  as  part  of  every  experiment, 
not  a  separate  exercise.  Whether  or  not  an  experiment  is  going  to  suc- 

iii 


iv  PREFACE 

ceed  has  no,  bearing  on  the  recording  of  the  notes,  for  the  notes  on  a 
failing  experiment  may  be  more  valuable  than  those  on  a  successful  one. 

The  teacher's  attention  is  called  to  the  Appendix  of  this  manual, 
where  suggestions  are  given  regarding  the  organization  of  the  laboratory 
work,  and  regarding  apparatus,  chemicals,  etc.  These  suggestions  are 
intended  simply  to  give  the  teacher  who  is  just  taking  up  the  course 
the  benefit  of  the  author's  experience,  and  thus  considerably  lighten  the 
burden. 

W.   H.   C. 

OBERLIN,  OHIO, 

March  10.  1922. 


TABLE   OF  CONTENTS 


CHAPTER  PAGE 

I.  KINETIC  THEORY.- 

Exp .    1 .  Brownian  Movement 1 

Exp.    2.  Air  Pressure  and  Rate  of  Evaporation 1 

II.  THE  GAS  LAWS. 

Exp.    3.  Boyle's  Law 3 

Exp.    4.  Partial  Volumes  and  Pressures  of  the  Gases  in  the  Air 5 

Exp.    5.  The  Coefficient  of  Expansion  for  Air 6 

Exp.    6.  Graham's  Law  of  Inverse  Proportionality 8 

Exp.    7.  Graham's  Law  and  Molecular  Weights 9 

III.  LAWS  GOVERNING  CHANGE  OF  STATE. 

Exp.    8.  Vapor  Pressure  of  Water 12 

Exp.    9.  Heat  of  Vaporization  of  Water 15 

IV.  MOLECULAR  WEIGHTS. 

Exp.  10.  Molecular  Weight  of  Carbon  Dioxide 17 

Exp.  11.  Molecular  Weight  of  Ether 19 

V.  THE  LAWS  OF  COMBINATION. 

Exp.  12.  The  Composition  of  Silver  Oxide 21 

Exp.  13.  Composition  of  Silver  Chloride 24 

Exp.  14.  Multiple  Proportions  of  Chlorine  in  the  Chlorides  of  Mercury  26 

Exp.  15.  The  Law  of  Volumes  from  the  Analysis  of  Ammonia 27 

VI.  ATOMIC  WEIGHTS. 

Exp.  16.  Specific  Heat  and  Atomic  Weight  of  Tin 30 

VII.  VALENCE. 

Exp.  17.  Valence  of  Sodium,  Magnesium,  and  Aluminum 32 

Exp.  18.  Oxidation  and  Reduction  Valence 34 

Exp.  19.  Zinc  as  a  Reducing  Agent 38 

VIII.  SOLUBILITY  AND  SUPERSATURATION  :  CONCENTRATION. 

Exp.  20.  Supersaturated  Solutions  of  the  Hydrates  of  Sodium  Sul- 
phate   40 

Exp.  21.  The  Test  for  Potassium 41 

Exp.  22.  Normal   Solutions   of  Hydrochloric,    Nitric,   and  Sulphuric 

Acids 41 

Exp.  23.  Normal  Sodium  Hydroxide 44 

v 


vi  TABLE  OF  CONTENTS 

CHAPTER  PAGE 

IX.  FREEZING    POINTS    AND    BOILING    POINTS    OF    SOLUTIONS:    OSMOTIC 
PRESSURE. 

Exp.  24.  Molecular  Lowering  of  the  Freezing  Point  of  Water 45 

Exp.  25.  Molecular  Weight  of  Propyl  Alcohol 47 

Exp.  26.  Qualitative  Experiment  on  Osmotic  Pressure 47 

X.  THEORY  OF  IONIZATION. 

Exp.  27.  Salt  Effect 48 

Exp.  28.  Degree  of  lonization  from  Abnormal  Freezing-point  Lowering.  48 

Exp.  29.  lonization  and  Chemical  Tests 49 

Exp.  30.  lonization  and  Catalysis 50 

Exp.  31.  Heat  of  Neutralization 52 

XI.  INDICATORS. 

Exp.  32.  Sensitiveness  of  Methyl  Orange  Indicator  and  Its  End-point 

Correction 54 

Exp.  33.  Sensitiveness  of  Phenolphthalein  Indicator 55 

Exp.  34.  Choice  of  an  Indicator 56 

Exp.  35.  Titration  of  Polybasic  Acids 57 

XII.  HOMOGENEOUS  EQUILIBRIUM. 

Exp.  36.  Speed  of  Reaction  and  Speed  Constant 59 

Exp.  37.  Equilibrium  Constant 61 

Exp.  38.  Ionic  Equilibrium  of  Cupric  Bromide 63 

Exp.  39.  Common-ion  Effect  with  Acetic  Acid  and  Ammonium  Hy- 
droxide    64 

Exp.  40.  Neutralization  by  Formation  of  a  Non-ionized  Acid  or  Base .  .  65, 

Exp.  41.  Hydrolysis  of  Salts 66 

XIII.  HETEROGENEOUS  EQUILIBRIUM. 

Exp.  42.  Decomposition  of  the  Hydrates  of  Cupric  Sulphate 69 

Exp.  43.  Partition  of  Bromine  between  Water  and  Carbon  Tetra- 

chloride 70 

Exp.  44.  Partition  of  Succinic  Acid  between  Water  and  Ether 71 

Exp.  45.  The  Cooling  Curve  of  Sodium  Sulphate  in  Light  of  the  Phase 

Rule 71 

XIV.  COMPLEX  EQUILIBRIUM. 

Exp.  46.  Precipitation  and  Solution  of  Silver  Acetate 74 

Exp.  47.  Precipitation  by  Means  of  Hydrogen  Sulphide  and  Its  Salts.  .  75 
Exp.  48.  Precipitation    of    Magnesium    by    Means    of    Ammonium 

Hydroxide 76 

Exp.  49.  The  Silver-Ammonium  Complex 77 

Exp.  50.  The  Ferric-Oxalate  Complex 79 

Exp.  51.  Study  of  the  Aluminum  Group 80 

Exp.  52.  Amphoteric  Nature  of  the  Halogens 81 


TABLE  OF  CONTENTS  Vll 

CHAPTER  PAGE 

XV.  ELECTROCHEMISTRY. 

Exp.  53.  Determination  of  the  Faraday 83 

Exp.  54.  Electrode  Reactions 84 

Exp.  55.  Migration  Velocity  of  Hydrogen  and  Hydroxyl  Ions 85 

Exp.  56.  Migration  of  a  Complex  Ion 87 

Exp.  57.  The  Daniell  Cell 87 

Exp.  58.  A  Concentration  Cell 89 

Exp.  59.  Decomposition  Voltage 89 

Exp.  60.  Displacement  Reactions 90 

APPENDICES 

I.  General  Outline  of  Laboratory  Work 93 

II.  Grouping  of  Students  for  the  Use  of  Special  Apparatus 94 

III.  Data  and  Suggestions  Regarding  Individual  Experiments. 96 

IV.  Chemicals 107 

V.  Apparatus Ill 

VI.  Table  of  Logarithms 114 

VII.  Table  of  Atomic  Weights Inside  back  cover 


LABORATORY   MANUAL 


CHAPTER  I 
KINETIC  THEORY 

Exp.  1.  Brownian  Movement. 

Apparatus. — A  compound  microscope,  a  slide  and  a  cover  glass. 

Procedure. — Take  about  0.01  gm.  (do  not  weigh)  of  gum  gamboge, 
and  rub  in  the  palm  of  the  hand  with  a  drop  or  two  of  water  to  a 
smooth  paste.  Dilute  somewhat,  place  a  drop  of  the  suspension  on  the 
slide  and  then  put  the  cover  glass  in  place.  With  a  10  X  eyepiece  and  a 
4  mm.  objective,  there  will  be  no  difficulty  in  seeing  that  some  of  the 
more  or  less  isolated  particles  are  in  rapid  vibration. 

You  should  apply  to  the  instructor  for  directions  in  using  the  instru- 
ment. Be  very  careful  in  any  case  not  to  turn  the  objective  down 
against  the  cover  glass,  or  injury  to  the  instrument  may  result. 

Write  out  this  experiment  in  your  notebook,  describing  exactly 
what  you  see.  Explain  just  why  the  particles  are  moving.  Why  do 
the  smallest  particles  move  most?  Do  any  of  the  particles  take  on  a 
rotary  motion?  Is  the  motion  here  seen  a  proof  of  the  existence  of 
molecules,  or  merely  evidence? 

Exp.  2.  Air  Pressure  and  Rate  of  Evaporation. 

Apparatus. — Two  desiccators  of  about  the  same  size  and  shape, 
one  of  them  connected  with  a  manometer  and  air  pump,  as  seen  in 
Fig.  1 ;  a  good  air  pump,  preferably  a  mechanical  pump  run  by  a  motor. 


To  Pump 


FIG.  1. 


2  KINETIC  THEORY 

If  a  water  pump  must  be  used,  the  process  of  exhausting  the  air  must 
not  be  stopped  by  turning  off  the  water,  or  the  latter  will  be  drawn  back 
into  the  manometer.  Close  cocks  A  and  B  (see  sketch),  and  then  pull  off 
the  rubber  tube  attached  at  C.  The  cock  B  may  then  be  opened  very 
cautiously,  whereupon  the  mercury  will  be  forced  up  into  the  end  of  the 
manometer  tube.  If  the  cock  is  opened  too  quickly  the  mercury  will 
go  over  with  a  thud,  probably  breaking  the  tube. 

Practically  the  same  procedure  must  be  followed  where  a  motor 
pump  is  used.  The  cocks  should  be  closed  before  the  pump  is  turned 
off,  and  it  is  also  well  to  have  on  the  rubber  tube  near  C  a  screw  clamp, 
or  stop-cock.  This  should  also  be  closed  before  the  pump  is  turned  off, 
or  oil  may  be  thrown  back  into  the  manometer. 

The  easiest  way  to  break  the  vacuum  is  cautiously  to  withdraw  the 
stop-cock  A,  thus  slowly  admitting  the  air. 

Procedure. — Place  in  a  watch  glass  in  each  desiccator  exactly  1  cc, 
of  distilled  water  measured  with  a  pipette.  Close  the  desiccators,  record 
the  time,  and  then  in  the  one  case  reduce  the  air  pressure  to  about  2  cm. 
of  mercury.  When  this  is  done,  close  the  cock  A  and  then  turn  off  the 
pump  as  directed  above.  The  two  desiccators  may  now  be  left  undis- 
turbed. 

After  waiting  about  ten  minutes  it  will  be  well  to  find  whether  the 
desiccator  is  "  holding  the  vacuum."  Proceed  as  follows:  Leave  the 
cock  A  closed,  run  the  pump,  if  necessary,  until  the  manometer  again 
registers  2  cm.,  and  then  cautiously  open  A.  The  manometer  should 
not  show  an  increase  in  pressure. 

Note  the  process  of  evaporation  in  the  two  cases,  and  when  it  is  com- 
pleted in  one  case  remove  the  other  watch  glass  and  weigh  to  determine 
what  weight  of  water  has  evaporated  from  it.  From  the  different 
weights  of  water  evaporated  in  equal  time  calculate  the  relative  rates. 

Explain  fully  what  you  have  observed. 


CHAPTER  II 


THE  GAS  LAWS 

Exp.  3.  Boyle's  Law. 

Apparatus.  —  The  Boyle's  law  apparatus  as  seen  in  Fig.  2.  Two 
glass  tubes  about  1  meter  long  and  6  mm.  bore  project  upward 
from  the  mercury  cistern  A.  One  is  closed  at  the  top  and  contains  the 
air  whose  volume  and  pressure  are 
to  be  measured.  The  scale  beside 
this  tube  indicates  the  volume.  The 
other  tube  is  open  at  the  top  and 
is  intended  as  a  manometer  to  indi- 
cate  pressures.  By  means  of  a  small 
bicycle  pump  attached  through  the 
stop-cock  tube  B  air  may  be  forced 
into  the  cistern  so  as  to  cause  the 
mercury  to  rise  in  both  the  upright 
tubes.  When  the  mercury  stands 
at  the  same  level  in  both  tubes  the 
pressure  in  the  tubes  is  the  same,  and 
is  equal  to  that  of  the  atmosphere 
outside.  When  the  mercury  stands 
higher  in  the  open  tube  the  pressure  ,  — 
in  the  closed  tube  is  equal  to  that  of  [ 
the  atmosphere  plus  the  differences 
in  the  levels  of  the  mercury.  This 

difference  in  levels  is  read  off  by  means  of  a  sliding  meter  scale  seen  in 
the  sketch  (C). 

The  tube  leading  from  the  pump  contains  a  valve  which  prevents 
the  recession  of  the  mercury  between  strokes.  This  valve  is  not  tight 
enough  to  depend  on  while  making  measurements:  hence  the  stop- 
cock B.  This  stop-cock  should  be  fastened  in  place  by  means  of  a 
rubber  band  to  prevent  leakage. 

Procedure.  —  Pump  up  the  apparatus  so  that  the  mercury  in  the 
manometer  stands  near  the  top  of  the  tube.  In  doing  this  take  short, 
quick  strokes,  and  be  very  careful  not  to  force  the  mercury  out  the  top 
of  the  tube.  When  the  pressure  is  sufficiently  high  catch  the  mercury 

3 


FIG.  2. 


4  THE  GAS  LAWS 

in  position  by  quickly  closing  the  stop-cock.  After  waiting  a  few 
moments  to  allow  the  vibration  of  the  mercury  column  to  cease  and  to  be 
sure  that  there  is  no  leakage,  proceed  to  measure  the  volume  and  pres- 
sure. The  volume,  V,  may  be  taken  directly  from  the  scale  to  the  left, 
and  should  be  read  to  hundredths  of  a  cubic  centimeter.  To  obtain 
the  pressure,  first  read  the  barometer  (in  centimeters  and  tenths) 
and  then  adjust  the  sliding  scale  so  that  the  metal  gauge  D  at  the  end 
coincides  exactly  with  the  top  of  the  mercury  in  the  closed  tube,  and 
read  off  the  difference  in  levels  to  0.1  cm.  The  pressure  P  is  the  sum 
of  the  barometric  pressure  and  the  difference  in  levels.  Calculate  the 
value  of  PV. 

Having  determined  one  value  for  PV,  cautiously  open  the  cock  B 
enough  to  allow  the  pressure  to  drop  somewhat,  say  10  cm.,  and  then 
redetermine  PV  as  directed  above.  This  may  be  called  P'V,  and 
should  not  vary  from  PV  by  more  than  two  or  three  units  out  of  a  total 
of,  say,  1750.  This  simply  means  that  PV  is  a  constant  for  any  given 
quantity  of  gas,  provided  the  temperature  remains  constant,  as  in  this 
case. 

Proceed  in  the  same  way  to  determine  other  values  for  PV  until 
six  or  seven  have  been  taken,  Record  the  results  thus : 


F. 

Differences 
in  Levels. 

Barometer. 

P. 

PV. 

After  having  worked  out  the  values  for  PV  as  above,  obtain  a  sheet 
of  centimeter  coordinate  paper  and  plot  the  values  for  P  and  V.  Finally 
by  use  of  a  "  French  curve,"  draw  a  smooth  line  through  the  plotted 
points,  obtaining  thus  the  curve  for  Boyle's  law.  Since  the  values  for 
P  are  much  larger  numerically  than  those  for  V  it  will  be  best  to  plot 
them  to  a  larger  scale . 

Find  the  ratio  between  any  two  pressures  and  then  find  the  inverse 
ratio  between  the  corresponding  volumes.  Are  they  the  same  within 
the  limits  of  experimental  error?  Are  we  right,  then,  in  saying  that  the 
volume  of  a  gas  varies  inversely  as  the  pressure? 

What  name  is  applied  to  the  curve  obtained  by  plotting  the  Boyle's 
law  data? 


PARTIAL  VOLUMES  AND  PRESSURES 


FIG.  3. 


Exp.  4.  Partial  Volumes  and  Pressures  of  the  Gases  in  the  Air. 

Apparatus. — The  apparatus  is  such  as  is  used  in  gas  analysis  for  the 
absorption  of  the  constituents  of  illuminating  gas  (Fig.  3) .  The  tube  A  is 
called  the  "  measuring  tube."  Tube  B  is  the  "  leveling  tube."  The 
absorption  pipette  C  con- 
tains metallic  copper  sur-  _  ft  ]\D 
rounded  with  a  solution  of 
ammonium  chloride  and 
ammonia.  In  the  presence 
of  oxygen  copper  goes  into 
solution  to  form  the  blue 
complex,  Cu(NH3)4(OH)2. 
Oxygen  is  thus  rapidly  ab- 
sorbed. The  solution  in  the 
spherical  bulbs  serves  to 
protect  that  in  the  pi- 
pette C. 

Procedure . — Disconnect 

the  capillary  tube  at  D,  and    

then  raise  B  until  the  water 

flows   over   and   expels   all 

the  gas  from  the  measuring  tube  A.     Now  lower  B  until  the  water  in 

A  stands  at  the  100-cc.  mark,   and  that  in  B  is  at  the   same   level. 

While   holding  the  tubes  in  this  position  let  another   person   close   the 

pinch  cock  E.     A  now  contains  100  cc.  of  moist  air  at  atmospheric 

pressure. 

First  be  sure  that  the  absorption  pipette  C  is  entirely  filled  with  liquid; 
then  connect  the  capillary  tube  securely  with  D  again,  remove  the  pinch 
cock,  and  raise  B  until  all  the  air  goes  over.  With  the  tubes  in  this 
position,  again  put  the  pinch  cock  securely  in  place.  Leave  the  appa- 
ratus now  for  ten  minutes,  shaking  the  absorption  pipette  occasionally 
to  hasten  the  absorption  of  the  oxygen.  Finally  run  all  the  residual  gas 
back  into  the  measuring  tube,  and  again  close  the  pinch  cock.  After 
adjusting  levels  read  the  volume  accurately. 

After  recording  the  volume,  repeat  the  process  of  absorption  to  make 
sure  that  all  the  oxygen  is  removed.  If  the  second  reading  of  the  vol- 
ume agrees  with  the  first  the  process  may  be  considered  finished.  Record 
the  final  volume. 

The  gas  left  in  the  tube  is  mostly  nitrogen,  but  contains  some  argon, 
and  some  water  vapor,  with  traces  of  other  gases.  Let  us  first  calculate 
the  pressure  which  this  mixture  exerted  when  mixed  with  the  oxygen. 
To  this  end  find  the  atmospheric  pressure  by  reading  the  barometer. 


6 


THE  GAS  LAWS 


This  was  the  total  pressure  including  the  sum  of  the  partial  pressures  of 
all  the  gases.  The  nitrogen  mixture  left  in  the  tube  at  the  end  of  the 
experiment  made  up  the  same  fraction  of  the  total  pressure  as  it  did  of 
the  total  volume,  and  what  this  was  you  already  have  the  data  to 
determine. 

Having  determined  the  pressure  of  the  residual  mixture,  refer  to 
the  table  in  the  manual,  page  14,  and  obtain  the  pressure  of  the  water 
vapor  for  the  corresponding  temperature.  Determine  then  the  pressure 
of  the  nitrogen  mixture  without  the  water  vapor.  Counting  the  pressure 
of  the  argon  as  0.8  per  cent  of  the  total  pressure  and  ignoring  the  trace 
of  other  gases,  calculate  the  pressure  of  the  nitrogen  alone. 

Now  calculate  the  volume  of  the  oxygen  removed  and  then  its 
pressure.  This  is  done  as  in  the  case  of  the  nitrogen  mixture. 

As  a  final  summary  of  the  experiment,  write  down  the  partial  pres- 
sures of  all  the  gases  in  the  air  as  you  have  found  them,  and  opposite  each 
place  its  partial  volume,  that  is,  the  number  of  cubic  centimeters  each 
would  occupy  at  atmospheric  pressure  if  removed  as  oxygen  was. 


Pressure. 

Volume. 

Nitrogen 

Oxygen  
Argon  
Water  vapor 

Other  gases  

trace 

trace 

Totals 

(The  partial  pressures  should  total  the  barometric  pressure  and  the 
partial  volumes  should  total  100  cc.) 

Exp.  5.  The  Coefficient  of  Expansion  for  Air. 

Apparatus. — The  most  important  part  of  the  apparatus  is  a  capillary 
tube  closed  at  one  end  and  containing  a  mercury  index.  The  tube 
must  have  a  uniform  bore  of  about  1  mm.  Select  a  tube  with  apparently 
uniform  bore,  and  test  by  drawing  in  a  column  of  mercury  about  10  cm. 
long,  and  then  measuring  this  column  when  in  different  positions.  If 
the  bore  is  uniform  the  column  of  mercury  will  have  the  same  length 
wherever  placed.  Cut  the  tube  40  cm.  long.  To  remove  any  possible 
traces  of  moisture  from  the  inner  surface  of  the  tube  force  through  it  a 
gentle  current  of  dry  air  while  heating  cautiously  with  a  Bunsen  flame. 

When  the  tube  is  cool  enough  to  handle,  seal  off  bluntly  10  cm.  from 


THE  COEFFICIENT  OF  EXPANSION  FOR  AIR 


one  end.  For  introduction  of  the  index  draw  out  a  piece  of  tubing 
until  it  is  barely  small  enough  to  pass  down  into  the  capillary.  If  this 
tube  is  thrust  into  a  deep  layer  of  mercury  and  closed  with  the  thumb, 
enough  mercury  may  be  withdrawn  to  form  the  index.  This  is  then 
introduced  by  placing  the  end  of  the  tube  about  10  cm.  down  the  capillary 
and  removing  the  thumb. 

Whenever  the  capillary  needs  drying  out,  the  index  may  be  thrown 
out  and  the  tube  heated  while  a  current 
of  air  is  drawn  through  it  by  means  of  a 
slender  tube  extending  to  the  bottom.  A 
new  index  is  then  introduced  as  above 
described. 

It  is  a  good  thing  to  keep  the  tube 
connected  with  a  calcium  chloride  drying 
tube  when  not  in  use.  If  no  precaution 
is  taken  moisture  will  work  in  past  the 
index,  alternately  condensing  and  vapor- 
izing when  the  tube  is  used  and  thus  in- 
troducing an  error  into  the  results. 

The  rest  of  the  apparatus  needs  little 
description.  B  is  a  so-called  Kjeldahl 
flask  used  in  nitrogen  determinations.  It 
stands  on  the  asbestos  ring  C  during 
heating,  to  prevent  superheating  of  the  steam. 

Procedure. — The  length  of  the  air  column  in  the  closed  end  of  the 
tube  is  measured  at  two  temperatures,  and  the  expansion  per  degree  is 
then  calculated  in  terms  of  the  volume  at  0°  C. 

Arrange  the  apparatus  first  as  seen  in  A,  adjusting  the  bottom  of  the 
tube  in  the  end  of  a  cork  so  that  the  surface  of  the  cork  coincides  with  the 
end  of  the  capillary,  and  holding  the  tube  securely  in  place  by  means  of  a 
ring-stand.  While  the  tube  is  in  this  vertical  position  accurately 
measure  the  length  of  the  column  of  air  below  the  index,  tapping  with  a 
pencil  to  prevent  any  lag  of  the  mercury,  and  taking  care  not  to  warm  the 
tube  or  the  thermometer  with  the  hands.  This  length  represents  V\. 
Also  read  the  thermometer  as  accurately  as  possible  (note  1).  This 
reading  is  t\. 

Push  the  stopper  a  down  the  tube  until  its  upper  surface  exactly 
coincides  with  the  top  of  the  index  and  then  measure  the  distance  down 
from  the  top  of  the  tube. 

Now  connect  the  apparatus  as  seen  in  B,  taking  care  that  the 
thermometer  and  tube  both  clear  the  sides  of  the  flask,  and  then  heat 
the  water  to  boiling,  applying  a  small  flame  directly  to  the  flask  without 


FIG.  4. 


8  THE  GAS  LAWS 

a  gauze.  The  index  will  rise,  of  course,  because  of  the  expansion  of  the 
air  below  it.  As  it  does  this  keep  pushing  the  tube  down  through  the 
stopper  so  that  the  upper  edge  of  the  index  coincides  with  its  upper  sur- 
face, thus  making  sure  that  all  the  air  column  is  heated.  Finally,  when 
the  index  no  longer  rises  even  after  tapping  with  a  pencil,  again  carefully 
measure  the  distance  down  from  the  top  of  the  tube.  The  difference 
between  this  measurement  and  the  former  one  will  show  the  amount  of 
expansion,  and  V2  is,  of  course,  equal  to  V  i  plus  this  expansion.  Record 
the  temperature  as  t^. 

When  the  data  have  been  obtained,  the  calculation  proceeds  as 
follows:  Let  x  equal  the  coefficient  of  expansion  for  air,  that  is,  the 
fraction  of  its  own  volume  at  0°  C.  which  it  expands  for  each  degree 
through  which  it  is  heated.  Let  1  equal  the  volume  at  0°  C.  At  ti  the 
volume  will  be  1  -\-t\x,  and  at  ti  the  volume  will  be  \-\-tix.  From  this  we 
obtain: 

Vi  :  V2''l+tix:  1+V; 

Substituting  your  data  in  this  equation,  you  can  at  once  work  out  the 
value  for  x. 

After  thus  obtaining  the  value  as  a  decimal,  transform  it  into  a  com- 
mon fraction.  The  standard  value  for  air  is  0.003666  or  1/272.8. 

NOTE. — For  accurate  work  the  thermometer  should  have  been  checked  by  com- 
parison with  an  accurate  standard  thermometer. 

Exp.  6.  Graham's  Law  of  Inverse  Proportionality. 

Apparatus. — A  glass  tube  1  meter  long  and  of  2  cm.  bore,  with  a 

tight-fitting  cork  for  each  end 

^^  ^  (Fig.  5).    Two  porcelain  boats  of 

FIG.  5.  equal  size  and  small  enough  to 

pass  easily  into  the  ends  of  the 

tube.  The  tube  should  be  suspended  in  a  horizontal  position  by  means 
of  two-ring  stands. 

Procedure. — Place  in  one  of  the  boats  a  few  cc.  of  concentrated 
hydrochloric  acid  and  in  the  other  a  like  amount  of  concentrated  ammo- 
nium hydroxide  (note  1).  When  the  two  boats  are  ready,  place  them 
in  the  two  ends  of  the  tube  at  the  same  time  (this  will  require  two 
people),  and  then  immediately  insert  the  corks.  In  about  fifteen  min- 
utes the  gaseous  ammonia  and  hydrogen  chloride  will  have  diffused 
along  the  tube  until  they  meet  somewhere  between  the  two  boats.  The 
point  of  contact  will  be  marked  by  a  sharply  defined,  disk-like  cloud  of 
ammonium  chloride.  Mark  on  the  tube  the  point  where  this  meeting 
occurs. 

Now   calculate  where  the  meeting  would  occur  if  the  distances 


GRAHAM'S  LAW  AND  MOLECULAR  WEIGHTS 


traveled  by  the  two  gases  were  inversely  proportional  to  the  square  roots 
of  their  densities,*  and  then  compare  with  the  observed  values. 

After  making  the  above  comparison,  allow  the  process  of  diffusion 
to  continue  for  some  time.  A  band  of  solid  ammonium  chloride  will  be 
formed.  Note  the  direction  in  which  the  band  broadens  and  explain. 

NOTE.  —  To  make  a  fair  comparison  of  the  rates  of  diffusion  the  gases  must 
be  of  the  same  molecular  concentration.  This  is  approximately  the  case  when  the 
solutions  are  of  the  concentrations  recommended. 

Exp.  7.  Graham's  Law  and  Molecular  Weights. 

Apparatus.  —  The  glass  cylinder  A  (Fig.  6)  should  be  about  40  cm.  high, 
and  4  cm.  in  diameter.  B  is  an  unglazed  battery  cup,  holding  about  80  cc. 
It  is  supported  by  a  rubber  stopper 
and  tube  as  seen.  The  rest  of  the 
apparatus  hardly  needs  description. 
The  amount  of  water  in  the  bottle 
C  should  be  somewhat  more  than 

sufficient  to  fill  the  cylinder  up  to  ^_=j  ^tQJ^JL-  E 

the  stopper,  but  not  enough  to  rise 
into  the  porous  cup.  M  and  N  are 
paper  markers. 

Procedure  A.  —  Remove  the 
pinch-cock  D  entirely,  so  as  to 
allow  free  passage  of  the  water, 
and  then,  by  means  of  a  pump  or 
otherwise,  apply  a  gentle  suction 
at  F  until  all  the  water  in  the  cyl- 
inder has  been  displaced.  Now 
disconnect  the  pump,  close  E,  and 
allow  the  air  to  diffuse  out  through 
the  porous  cup,  noting,  by  means 
of  a  stop-watch  (or  clock),  the 
exact  time  required  for  the  surface 
of  the  water  in  the  cylinder  to  pass 
from  one  marker  to  the  other.  Repeat  the  determination,  obtaining 
two  concordant  values.  The  average  of  these  values  may  be  recorded 
as  t  (note  1).  The  average  molecular  weight  of  air  is  29.  This  is  to  be 
used  as  your  standard  of  comparison  in  determining  the  molecular 
weight  of  other  gases. 

*  Prepare  two  simultaneous  equations,  thus: 


FIG.  6. 


x  +y  =  length  of  tube. 


10  THE  GAS  LAWS 

Procedure  B. — Substitute  carbon  dioxide  for  air,  running  in  the  gas 
from  a  cylinder  or  generator,  not  too  rapidly,  and  applying  suction  at 
F  if  necessary.  After  the  cylinder  is  filled,  the  gas  should  be  allowed  to 
bubble  through  the  water  in  the  bottle  C  for  several  minutes.  If  this 
precaution  is  not  taken,  solution  of  the  gas  in  the  water  will  take  place 
during  the  process  of  diffusion,  and  the  time  will  thus  be  shortened. 
Having  properly  filled  the  cylinder,  determine  the  time  of  diffusion  as 
was  done  with  air,  obtaining  two  concordant  values,  and  taking  the 
average  (£')• 

Calculating  from  your  data,  determine  whether  the  times  required 
for  the  diffusion  of  equal  volumes  of  carbon  dioxide  and  air  are  pro- 
portional to  the  square  roots  of  the  molecular  weights,  in  other  words, 
whether  according  to  your  data  the  proportion 

t  :  t'::Vm  ;  Vm' 

actually  holds  good. 

Assume  also  that  the  molecular  weight  of  carbon  dioxide  is  unknown, 
and  calculate  its  value  from  your  data,  using  the  equation 

t'2Xm 
m'.'— 

m  in  this  case  being  29,  the  molecular  weight  of  air. 

Procedure  C. — Determine  the  molecular  weight  of  the  natural  gas 
used  in  your  burners.  Do  not  forget  the  precaution  concerning  solu- 
bility. 

Natural  gas  is  usually  nearly  pure  methane,  CH4,  having  a  molecular 
weight  of  16.  Some  samples,  however,  contain  gasoline  vapor.  Gaso- 
line is  a  mixture  of  hydrocarbons  having  an  average  molecular  weight  of 
about  100.  Consequently,  samples  of  gas  containing  this  will  have  a 
higher  molecular  weight  than  pure  methane.  When  the  molecular 
weight  is  above  19  the  gas  is  usually  considered  "  rich  "  enough  to  work 
over  for  the  gasoline  it  contains. 

After  determining  the  molecular  weight  of  the  gas,  decide  whether 
it  would  be  profitable  to  work  it  over  in  this  way. 

Procedure  D. — Determine  the  molecular  weight  of  hydrogen, 
(note  2),  proceeding  as  in  the  above  cases.  Hydrogen  is  not  appre- 
ciably, soluble  in  water,  but  the  precaution  about  solubility  must  be 
observed,  or  the  hydrogen  will  be  mixed  with  a  considerable  amount  of 
the  gas  last  used. 

In  all  these  diffusion  experiments  it  must  be  remembered  that  the 
gases  are  saturated  with  water  vapor.  In  cases  where  the  molecular 
weight  of  the  gas  is  equal  to,  or  near,  that  of  water,  this  has  no  appre- 


GRAHAM'S  LAW  AND  MOLECULAR  WEIGHTS  11 

ciable  effect  on  the  rate  of  diffusion;  but  in  the  case  of  hydrogen,  a  gas 
of  very  low  molecular  weight,  the  effect  is  very  marked.  It  is  possible, 
however,  to  calculate  a  correction,  and  thus  get  a  fair  comparison  of 
values.  The  average  pressure  of  the  mixture  of  hydrogen  and  water 
vapor  during  diffusion  is  about  76  cm.  Of  this  pressure  about  74  cm. 
are  due  to  hydrogen  (mol.  wt.  2)  and  about  2  cm.  to  water  vapor 
(mol.  wt.  18).  Calculate  on  this  basis  the  average  molecular  weight 
of  the  mixture,  and  then  compare  with  the  experimental  value. 

NOTES. — (1)  The  time  for  air  should  be  about  ten  minutes. 

(2)  If  a  pressure  cylinder  is  used  a  pressure  regulator  must  be  attached. 


CHAPTER   III 
LAWS  GOVERNING  CHANGE  OF  STATE 

Exp.  3.  Vapor  Pressure  of  Water. 

A  known  volume  of  air  saturated  with  water  vapor  is  drawn  through 
a  calcium  chloride  tube  where  the  water  is  absorbed  and  weighed. 
From  the  weight  of  this  water  the  vapor  pressure  is  calculated. 

Apparatus. — The  sketch  (Fig.  7)  shows  the  general  arrangement.  The 
12-inch  tower  A  contains  wool  which  has  been  wet  and  then  squeezed 


FIG.  7. 

as  dry  as  possible.  It  should  be  packed  rather  loosely,  but  in  such  a 
way  as  not  to  leave  any  channels  through  which  the  air  may  pass  without 
coming  into  very  close  contact  with  it. 

The  5-inch  U-tube  B  contains  granulated  calcium  chloride  of  wheat- 
grain  size,  well  shaken  down.  At  the  top  of  each  limb  is  a  loose  plug 
of  asbestos  intended  to  prevent  the  expulsion  of  any  fine  dust  from  the 
surface. 

All  so-called  "  anhydrous  "  calcium  chloride  contains  some  water 
of  hydration,  and  its  effectiveness  as  a  drying  agent  depends  on  the 
amount  of  this  water  present.  It  also  usually  contains  some  calcium 
oxide,  which  absorbs  carbon  dioxide  from  the  air  and  thus  changes 
weight.  To  correct  both  of  these  defects,  the  tube,  after  being  filled,  is 

12 


VAPOR  PRESSURE  OF  WATER  13 

suspended  up  to  the  side  arms  through  a  slot  in  the  cover  of  an  air  bath, 
and  heated  to  275°  C.  while  a  slow  current  of  dry  carbon  dioxide  is 
conducted  through  it.  This  process  should  continife  for  at  least  an 
hour.  A  good  way  to  tell  when  the  water  is  all  removed  is  to  place  a 
cold  glass  tube  over  the  exit  arm  of  the  U-tube  and  note  whether  there 
is  any  condensation  of  moisture.  When  calcium  chloride  is  thus  treated 
it  becomes  almost  anhydrous,  and  the  surface  is  left  in  a  fine,  porous 
condition.  According  to  A.  T.  McPherson  *  the  reagent  thus  prepared 
takes  on  much  water  by  simple  adsorption  on  the  surface  of  the  par- 
ticles before  the  formation  of  a  hydrate  begins,  and  its  drying  capacity 
during  this  process  is  almost  as  perfect  at  that  of  P2O.5,  the  best  drying 
agent  known.  Its  efficiency  in  this  capacity  also  depends,  of  course, 
upon  the  length  of  the  column  and  upon  the  slowness  with  which  the 
air  is  drawn  through. 

At  the  end  of  the  drying  process  the  U-tube  is  disconnected  from 
the  CO2  generator,  and  while  it  is  still  hot,  dry  air  is  drawn  through  it 
to  displace  the  CO2.  (Why  is  this  necessary?)  Subsequent  drying  may 
be  done  with  air  alone. 

In  the  use  of  this  drying  tube  the  following  precautions  should  be 
observed : 

(1)  Keep  the  arms  of  the  tube  closed  when  not  in  use  to  prevent 
access  of  moisture.     (Rubber  connectors  containing  large  glass  beads 
are  good  for  this  purpose.) 

(2)  Keep  the  surface  of  the  tube  perfectly  clean  and  bright ;  other- 
wise the  weight  cannot  be  absolutely  depended  on. 

(3)  Always  run  the  current  of  air  in  the  same  direction.     If  the 
tube  is  used  in  one  direction  and  then  reversed,  much  of  the  water 
already  absorbed  will  be  expelled.     It  will  be  well  to  mark  the  stoppers 
"  1  "  and  "  2."  to  guard  against  this. 

The  thermometer  C  must  be  accurate  to  0.1°,  or  at  least  the  data  for 
a  correction  must  be  at  hand.  An  error  of  0.1°  in  reading  the  thermom- 
eter will  (at  20°)  produce  an  apparent  error  of  about  0.1  mm.  in  the 
vapor  pressure.  (Prove  this.) 

D  is  a  common  thermometer  which,  however,  should  be  correct  to  at 
least  1°  C. 

The  flask  E  is  graduated  to  hold  some  definite  amount,  say  4.5  liters; 
the  amount  should  not  be  much  less  than  this. 

It  is  scarcely  necessary  to  say  that  the  connections  must  be  secure. 

Those  attached  to  the  U-tube  must,  however,  be  easily  removable  without 

breaking  the  tube.     In  removing  a  connector  from  the  arm  of  a  U-tube, 

be  sure  to  grasp  the  tube  on  the  side  near  the  arm;    if  the  connector 

*Jour.  Am.  Chem.  Soc.,  July,  1917. 


14 


LAWS  GOVERNING  CHANGE  OF  STATE 


sticks,  roll  it  back  with  the  thumb,  twisting  gently  at  the  same  time. 
Do  not  pull  it  and  thus  cause  it  to  grip  the  tube  all  the  tighter. 

Procedure. — disconnect  the  U-tube  from  the  rest  of  the  apparatus, 
place  it  flat  upon  the  balance  pan,  and  weigh  accurately  to  the  fourth 
decimal  place.  To  be  sure  that  the  air  pressure  inside  the  tube  is  the 
same  as  that  of  the  atmosphere,  the  side  arms  may  be  left  open  while 
weighing.  Connect  the  weighed  U-tube  in  its  place  so  as  to  allow  the 
air  to  pass  in  the  proper  direction,  and  then  start  the  water  from  the 
aspirator  bottle  into  the  graduated  flask.  It  may  run  rapidly  enough 
to  form  a  barely  continuous  stream.  Record  the  temperature  as  read 
on  both  the  thermometers,  remembering  to  read  the  one  where  the  sat- 
uration occurs  to  0.1°,  and  correcting  the  reading  if  necessary. 

When  the  proper  volume  of  air  has  passed  through  the  apparatus, 
turn  off  the  aspirator  and  then  disconnect  the  U-tube  for  weighing. 
At  the  same  time  also  read  the  temperature  on  the  two  thermometers 
and  record  with  the  first  readings.  Finally  weigh  the  U-tube  as  before 
and  calculate  the  weight  of  the  water  collected. 

You  now  have  the  data  for  calculating  the  members  of  the  equation 
PV  =nRT,  and  when  this  is  done  the  value  for  P  can  at  once  be  deter- 
mined. Thus,  if  the  average  of  the  two  readings  on  thermometer  C  is 
within  1°  of  the  average  for  D  the  volume  of  water  run  off  represents 
the  volume,  V,  of  saturated  water  vapor  run  into  the  U-tube;  if  not, 
the  correct  volume  must  be  calculated  by  the  use  of  Charles'  law.  T 
is  the  average  of  the  two  readings  on  thermometer  C,  reduced,  of  course, 
to  the  absolute  scale;  n  is,  of  course,  the  fraction  of  a  mole  of  water 
collected. 

Calculate  P  by  use  of  the  equation  given  above,  remembering  that 
the  value  thus  obtained  will  be  some  fraction  of  an  atmosphere,  and 
finally  translate  the  value  into  millimeters.  Compare  this  value  with 
that  obtained  by  interpolation  from  the  table  below.  Your  value 
should  not  differ  from  this  bv  more  than  0.2  mm. 


VAPOR   PRESSURE   OF   WATER  IN   MILLIMETERS 


Temperature. 

Pressure. 

Temperature. 

Pressure. 

15 

12.7 

21 

18.5 

16 

13.5 

22 

19.7 

17 

14.4 

23 

20.9 

18 

15.4 

24 

22.2 

19 

16.3 

25 

23.6 

20 

17.4 

26 

25.1 

HEAT  OF  VAPORIZATION  OF  WATER 


15 


Exp.  9.  Heat  of  Vaporization  of  Water. 

Apparatus. — As  seen  in  Fig.  8.  The  Erlenmeyer  flask  A  is  of 
500-cc.  capacity.  The  trap  B  is  2.5  by  12  cm.  The  tube  E  serves  to 
draw  off  the  water  which  accumulates  from  condensation.  The  400-cc. 
calorimeter  beaker  C  stands  in  a  1-qt.  graniteware  cup,  and  is  insulated 
by  packing  about  loosely  with  cotton.  It  is 
covered  by  means  of  a  piece  of  cardboard 
cut  as  seen  at  F.  The  transite  board  D 
serves  to  protect  the  calorimeter  from  the 
heat  of  the  flame. 

The  apparatus  should  be  carefully  con- 
structed according  to  the  sketch,  and  offered 
for  inspection  before  it  is  used. 

Procedure.  —  Weigh  the  calorimeter 
beaker  accurately  on  the  laboratory  bal- 
ance, fill  about  two-thirds  full  of  distilled 
water  which  has  previously  been  cooled 
to  about  5°  C.  by  addition  of  ice  water 


FIG.  8. 


(note  1),  and  then  weigh  again.  In  the  meantime  heat  the  water  in 
the  Erlenmeyer  flask  to  rapid  boiling.  Now  pack  the  calorimeter  in 
the  cup  as  seen  in  the  sketch,  and  immediately  take  the  temperature 
to  0.1°,  stirring  well  while  doing  this.  Call  this  t\.  When  this  is  done 
immediately  thrust  the  steam  tube  into  the  water,  cover  the  calorimeter, 
and  then  allow  the  condensation  to  proceed  until  the  temperature  of  the 
water  is  as  much  above  room  temperature  as  the  original  temperature 
was  below  (note  2),  stirring  well  each  time  before  reading.  Now  quickly 
push  the  calorimeter  to  one  side  so  as  to  withdraw  the  steam  tube,  stir 
well,  and  immediately  take  the  temperature,  fe,  keeping  the  calorimeter 
covered  as  much  as  possible.  Finally  remove  the  calorimeter,  and 
weigh  as  at  first.  The  increase  in  weight  gives  the  amount  of  steam 
which  has  condensed. 

The  calculation  is  made  as  follows:  The  original  weight  of  water 
multiplied  by  the  change  in  temperature  (fe  —  ^i )  gives  the  heat  absorbed 
by  the  water.  But  the  glass  has  also  been  heated.  One  gm.  of  glass 
requires  0.2  calorie  to  raise  its  temperature  1°  C.  We  may  therefore 
calculate  the  heat  absorbed  by  the  glass  by  multiplying  its  weight  by 
0.2  and  then  by  the  change  in  temperature.  Or,  if  we  choose,  we  may 
calculate  the  water  equivalent  of  the  glass,  which  is  0.2  its  weight,  and 
this  may  be  added  to  the  weight  of  the  water  before  the  heat  absorbed 
by  the  latter  is  calculated.  At  any  rate  the  total  amount  of  heat 
absorbed  is  the  sum  of  the  amounts  absorbed  separately  by  the  water  and 
the  glass. 


16  LAWS  GOVERNING  CHANGE  OF  STATE 

Now  the  total  heat  absorbed  has  come  from  two  sources:  (1)  from 
the  condensation  of  the  steam  at  100°  and  (2)  from  the  cooling  of  the 
water  thus  condensed  from  100°  (note  3)  to  the  final  temperature,  fe. 
The  latter  must  be  calculated  by  multiplying  the  weight  of  the  water 
condensed  by  its  change  in  temperature  (100°  — fe).  If  we  subtract 
the  heat  so  liberated  from  the  total  heat  absorbed  we  shall  have  the 
heat  liberated  by  the  steam  in  condensing  at  100°. 

To  get  the  heat  liberated  by  the  condensation  of  1  gm.  of  steam  at 
100°  (the  heat  of  vaporization)  we  must  divide  the  amount  last  calcu- 
lated by  the  total  weight  of  water  condensed. 

NOTES. — (1)  Be  careful  to  see  that  no  pieces  of  ice  are  present  when  weighing. 
(Why?) 

(2)  By  beginning  below  room  temperature  and  ending  above  we  shall  have  a 
balancing  of  errors;   the  water  at  first  takes  up  heat  from  the  air  and  later  gives  it 
off. 

(3)  The  boiling-point  of  water  is  not  quite  100°  C.  at  our  average  pressure.     See 
"  Boiling-point  of  Water,"  Handbook  of  Chem.  and  Phys.*    The  value  here  found 
should  be  used  hi  the  calculation. 

*  Published  by  the  Chemical  Rubber  Co.,  Cleveland,  O. 


CHAPTER  IV 
MOLECULAR  WEIGHTS 

Exp.  10.  Molecular  Weight  of  Carbon  Dioxide. 

Apparatus. — A  250-cc.  Erlenmeyer  flask  fitted  with  tubes  and  pinch 
cocks  as  seen  in  Fig.  9.  The  external  ends  of  the  tubes,  and  the 
connectors  E  and  F  must  be  short,  or  difficulty  will  be  experienced  in 
getting  the  flask  upon  the  balance  pan.  Also  the  tube  connected  with 
E  must  not  extend  below  the  stopper.  There  should  be  a  mark  on  the 

From  C00 


generator 


FIG.  9. 

neck  of  the  flask  to  which  the  stopper  is  always  adjusted  in  order  that 
the  capacity  may  not  vary.  It  is  hardly  necessary  to  say  that  the  flask 
should  be  clean  and  dry  and  free  from  finger  marks  (note  1). 

The  other  pieces  of  apparatus  shown  in  the  sketch  are  a  gas  wash 
bottle  A  and  a  calcium  chloride  drying  column  B.  A  carbon  dioxide 
generator  or  a  pressure  cylinder  is  needed,  but  is  not  here  shown. 

The  thermometer  C  measures  the  temperature  of  the  gas. 

Procedure. — Take  the  clean,  dry  flask  prepared  as  above,  open 
one  of  the  pinch  cocks  so  that  the  pressure  inside  the  flask  shall  be  the 
same  as  that  of  the  atmosphere,  and  then  weigh  carefully  to  the  third 
decimal  place. 

Now,  from  a  generator  or  pressure  cylinder,  fill  the  flask  with  air- 
free  carbon  dioxide  (note  2) .  The  gas  should  pass,  first  through  a  solu- 

17 


18  MOLECULAR  WEIGHTS 

tion  of  sodium  bicarbonate  contained  in  the  wash  bottle  A  (note  3), 
and  then  through  the  drying  column  B.  The  current  should  not  be  too 
rapid,  or  the  washing  and  drying  may  not  be  complete.  The  bubbles 
may  pass  about  as  rapidly  as  a  watch  ticks.  Let  the  gas  enter  the 
flask  through  the  longer  tube,  as  seen  in  the  sketch.  (Why?)  After  the 
process  has  continued  for  about  ten  minutes,  close  the  pinch  cocks,  and 
remove  the  flask  to  the  balance  for  weighing. 

After  weighing,  again  connect  the  flask  with  the  generator  as  before, 
and  pass  the  carbon  dioxide  through  it  for  another  five  minutes.  In  the 
meantime  take  the  temperature  and  barometric  pressure.  Finally, 
weigh  the  flask  again.  If  the  air  was  all  removed  on  the  first  trial  there 
will  be  no  change  in  weight.  (Why?)  A  very  slight  change  is  allow- 
able, say  1  mg. 

The  weight  of  the  carbon  dioxide  is  the  difference  between  the 
weight  of  the  flask  full  of  the  gas  and  the  vacuous  flask,  and  its  volume  is, 
of  course,  the  volume  of  the  flask.  Both  the  weight  of  the  vacuous  flask 
and  its  capacity  are,  therefore,  to  be  determined. 

Determine  the  capacity  of  the  flask  by  filling  with  water  up  to  the 
pinch  cocks  and  weighing  on  the  laboratory  balance.  The  weight  of  the 
water,  and  so  its  volume  (note  4) ,  may  be  taken  as  the  difference  between 
the  weight  of  the  flask  in  air  and  the  above  weight.  To  find  the  weight 
of  the  vacuous  flask  we  must  know  the  weight  of  air  it  contains  at  the 
observed  temperature  and  pressure.  This  may  best  be  determined 
from  the  table,  "  Density  of  Dry  Air,"  Handbook  of  Chem.  and  Phys., 
where  the  weight  of  1  cc.  of  air  for  any  ordinary  temperature  and 
pressure  is  given.  Knowing  the  weight  of  the  flask  filled  with  air  and 
the  weight  of  the  air  contained  in  it,  determine  the  weight  of  the  vacuous 
flask. 

You  now  have  all  the  necessary  data,  namely,  the  weight  of  the  flask 
filled  with  carbon  dioxide  gas,  the  weight  of  the  vacuous  flask,  and  the 
temperature  and  pressure  of  the  gas  when  collected.  If  you  determine 
the  weight  of  the  gas  and  then  reduce  its  volume  to  standard  conditions 
you  will  have  the  weight  of  a  known  volume  under  standard  conditions. 
You  then  have  only  to  find  the  weight  of  22.4  liters  under  the  same  con- 
ditions. This  will  be  the  molecular  weight.  For  the  sake  of  the  prac- 
tice, make  the  calculation  also  by  use  of  the  equation  PV  =wRT/M. 

NOTES. — (1)  "  Bon-ami  "  or  some  similar  preparation  is  good  for  cleaning  glass. 
When  an  article  is  properly  cleaned,  water  adhering  to  the  surface  will  not  creep  up 
into  drops  but  will  form  a  continuous  film.  A  solution  of  chromic  acid  in  sulphuric 
acid  (best  made  by  mixing  about  10  gm.  of  powdered  Na2Cr2O7  with  300  cc.  of  crude, 
concentrated  sulphuric  acid)  is  very  good  for  removing  traces  of  grease  from  the 
inside  of  apparatus  or  any  place  which  cannot  be  reached  with  other  cleaners.  This 
should,  however,  be  used  with  great  care,  due  to  its  very  corrosive  nature.  Never 


MOLECULAR  WEIGHT  OF  ETHER 


19 


leave  any  on  the  outside  of  the  bottle  to  get  upon  the  hands  or  clothing  or  to  spoil 
the  desk  top. 

Articles  cleaned  in  any  way  must  be  carefully  rinsed  first  with  tap  water  and  then 
with  distilled  water. 

To  dry  out  apparatus  after  cleaning,  first  throw  out  as  much  water  as  possible  by 
swinging  the  piece  violently  down  at  the  side,  then  heat  gently  over  a  gauze  (not  over 
a  direct  flame),  rotating  constantly,  and  at  the  same  time  conduct  into  the  piece  a 
gentle  stream  of  air  from  the  air-blast  line. 

(2)  The  carbon  dioxide  from  a  pressure  cylinder  is  not  likely  to  contain  much  air, 
but  that  from  a  generator  may  contain  a  considerable  amount,  depending  on  the 
length  of  time  the  generator  has  run  since  refilling. 

(3)  The  carbon  dioxide  is  likely  to  contain  a  little  hydrochloric  acid  gas  from  the 
generator.   (Why?)  By  interaction  with  sodium  bicarbonate  the  latter  is  absorbed,  an 
equivalent  amount  of  carbon  dioxide  being  set  free.     Why  would  normal  sodium 
carbonate,  Na2CO3,  not  do  just  as  well? 

(4)  One  gm.  of  water  occupies  a  volume  of  1  cc.  only  at  4°  C.,  and  this  means  1  gm. 
weighed  in  a  vacuum.     Weighed  in  air,  the  water  is  buoyed  up  somewhat,  and  so 
more  is  required  to  give  an  apparent  weight  of  1  gm.     At  20°  C.  the  volume  of  water 
corresponding  to  an  apparent  weight  of  1  gm.  is  1.003  cc.     The  difference  may,  how- 
ever, be  neglected  in  this  case. 

Exp.  11.  Molecular  Weight  of  Ether. 

Apparatus. — A  Dumas  bulb  in  which  the  end  of  the  sealing  tube  has 
been  replaced  by  a  rubber  connector  bearing  a  pinch  cock.  The  con- 
nector is  securely  fastened  in  place 
by  wiring,  and  the  end  of  the  wire  is 
turned  into  a  loop  by  which  the 
bulb  is  suspended  when  being 
weighed.  The  bulb  may  be  dried 
out  as  usual  (note  1,  Exp.  10),  the 
only  requisite  being  the  use  of  a 
very  slender  metal  tube  to  conduct 
the  air  in.  There  will  also  be  needed 
a  3-qt.  graniteware  bath,  a  bulb 
holder,  and  a  funnel  with  a  long, 
slender  stem  to  be  used  in  getting 
the  ether  into  the  bulb.  The  latter  may  be  made  by  drawing  out  the 
stem  of  a  thistle  tube. 

Procedure. — Determine  the  weight  of  the  bulb  filled  with  air  at  the 
temperature  and  pressure  of  the  balance  room;  that  is,  leave  the  bulb 
hanging  in  the  balance  case  with  the  pinch  cock  open  for  a  few  minutes 
before  weighing.  In  the  meantime  have  water  heating  in  the  bath  for 
the  submersion  of  the  bulb.  It  should  be  heated  to  about  90°  C. 

Put  about  10  cc.  of  pure,  dry  ether  in  the  bulb,  turn  out  all  flames  in 
the  vicinity,  and  then  by  means  of  the  holder  place  the  bulb  under  the 
hot  water  as  seen  in  the  sketch,  taking  pains  not  to  allow  the  outlet  tube 


FIG. 


20  MOLECULAR  WEIGHTS 

to  become  submerged.  Stir  the  bath  constantly  by  means  of  a  mechan- 
ical stirrer  or  by  slowly  moving  the  bulb  back  and  forth  in  a  lateral  direc- 
tion. Watch  for  the  ether  to  boil  away  completely,  and  after  this  has 
occurred  wait  about  two  minutes  for  the  vapor  to  gain  the  temperature 
of  the  bath  (note  1)  taking  care  not  to  raise  the  bulb  in  the  meantime 
(note  2).  Finally  take  the  temperature  of  the  bath  as  accurately  as 
possible,  close  the  pinch  cock  securely,  and  then  remove  the  bulb  and 
wipe  it  dry.  Hang  the  bulb  now  in  the  balance  case,  and  after  allowing 
ample  time  for  temperature  adjustment,  weigh.  In  the  meantime  read 
the  barometer. 

To  obtain  the  capacity  of  the  bulb,  fill  with  water  and  weigh.  The 
filling  is  easily  accomplished  if,  with  the  bulb  still  closed,  the  outlet  tube 
is  placed  under  water  and  the  pinch  cock  then  removed  (note  3).  If  the 
bulb  does  not  fill  completely  by  this  method,  it  is  due  to  the  presence 
of  air  which  was  not  expelled  when  the  ether  boiled  away  or  which  may 
have  leaked  in  during  the  weighing.  A  small  bubble  is  permissible. 
The  filling  must  be  complete  before  the  bulb  is  weighed,  of  course. 

Having  thus  determined  the  capacity,  determine  the  weight  of  the 
vacuous  bulb  as  directed  under  Exp.  10.  Calculate  then  the  weight  of 
the  ether  vapor.  You  then  have  the  weight  of  a  known  volume  of  gas- 
eous ether  at  a  known  temperature  (the  temperature  of  the  bath)  and 
at  a  known  pressure  (the  pressure  of  the  air).  Reduce  the  volume  to 
standard  conditions,  and  then  calculate  the  molecular  weight  in  the 
usual  way. 

NOTES. — (1)  While  the  ether  is  boiling  away  the  vapor  is  at  its  boiling-point, 
34°  C. 

(2)  If  the  flask  is  raised  after  the  liquid  is  all  boiled  away,  the  vapor  will  con- 
tract, and  air  will  enter. 

(3)  When  the  bulb  was  cooled  for  weighing,  much  of  the  ether  vapor  condensed 
to  a  liquid,  as  you  doubtless  noticed.     This,  of  course,  much  reduced  the  pressure, 
and  thus  made  possible  the  entrance  of  the  water.     The  ether  vapor  still  in  the  bulb 
dissolves  in  the  water,  still  further  lowering  the  pressure  and  thus  insuring  almost 
complete  filling  if  no  air  is  present. 


CHAPTER  V 
THE  LAWS  OF  COMBINATION 

Exp.  12. — The  Composition  of  Silver  Oxide. 

Apparatus. — An  air  bath  made  from  a  quart  graniteware  cup,  shaped 
as  seen  in  the  Fig.  11;  also  stirring  rods,  and  a  desiccator  as  shown  in 
Fig.  12. 

To  distribute  the  heat  more  evenly,  the  air  bath  contains  a  star- 
shaped  diaphragm  of  iron  or  tin.  The  points  of  this  are  bent  down  to 


f ^J~7C  .{;•£ '  iC-,J~-,~C^i\\/        * 


Gauza 


FIG.  11. 


FIG.  12. 


serve  as  legs,  thus  bringing  the  diaphragm  about  half  an  inch  above  the 
bottom  of  the  bath.  The  triangle  bearing  the  object  to  be  heated  stands 
astride  the  diaphragm,  but  high  enough  to  prevent  contact  of  the  object 
with  it.  The  thermometer  is  inserted  through  a  cork  cut  to  fit  the  spout 
of  the  cup,  and  projects  over  the  diaphragm  as  near  the  triangle  as  pos- 
sible. By  means  of  an  ordinary  Bunsen  flame  temperatures  up  to 
300°  C.  may  be  maintained.  In  using  the  bath  a  constant  temperature 
should  be  secured  before  the  object  to  be  heated  is  put  in.  For  low 
temperatures  it  will  be  best  to  use  a  small  luminous  flame,  which  is 
more  likely  to  be  steady. 

The  following  table,  showing  the  approximate  relationship  between 
size  of  flame  and  temperature,  will  be  helpful:  * 


*  Natural  gas  at  a  pressure  of  about  6  oz.  was  used. 
21 


22  THE  LAWS  OF  COMBINATION 

Size  and  Nature  of  Flame.  Temperature. 

1-inch,  luminous,  2  inches  from  bath 100°  C. 

2-inch,  luminous,  1  inch  from  bath 135°  C. 

2-inch,  non-luminous,  1  inch  from  bath 145°  C. 

3-inch,  non-luminous,  just  touching  bath 180°  C. 

4-inch,  non-luminous,  air  well  regulated 240°  C. 

5-inch,  non-luminous,  air  well  regulated 300°  C. 

Temperatures  higher  than  300°  C.  should  not  be  used,  or  the  enamel 
on  the  bath  will  be  melted. 

The  stirring  rods  should  be  about  17  cm.  long  and  not  more  than 
4  mm.  in  diameter.  The  ends  must  be  nicely  rounded  by  heating  in  a 
Bunsen  flame.  The  rods  must  in  no  case  be  used  without  this.  It  is 
well  to  have  one  rod  tipped  with  a  short  piece  of  small  rubber  tubing, 
which  should  be  allowed  to  project  slightly  over  the  end.  This  is  some- 
times indispensable  for  removing  the  last  traces  of  a  precipitate  from  a 
beaker. 

The  desiccator  should  be  carefully  cleaned,  both  inside  and 
outside.  If  it  already  contains  dry  calcium  chloride  this  may  be 
left  in.  If  not,  have  the  store-room  keeper  fill  the  bottom  part  about 
half  full,  pouring  in  the  calcium  chloride  through  a  funnel  so  as  not 
to  dust  the  upper  part  of  the  desiccator.  A  diaphragm  of  wire 
gauze  rests  on  the  shoulder  above  the  calcium  chloride,  and  upon  this 
should  stand  one  or  more  triangles  having  the  ends  bent  down  to  form 
legs  about  three-fourths  of  an  inch  high.  A  crucible  placed  in  a  desiccator 
to  cool  always  rests  on  the  triangle,  not  on  the  diaphragm.  Other 
objects,  for  example  a  weighing  bottle,  may  rest  directly  on  the  dia- 
phragm. The  cover  of  the  desiccator  is  made  air-tight  by  means  of 
a  very  thin  coat  of  vaseline.  An  excess  should  be  avoided. 

All  the  above  pieces  of  apparatus  should  be  in  order  before  any 
experimental  work  is  even  begun.  Ask  to  have  them  inspected. 

Procedure. — First  prepare  silver  oxide  as  follows :  Weigh  out  on  the 
laboratory  balance  about  4  gm.  of  silver  nitrate,  and  dissolve  this  in 
about  50  cc.  of  distilled  water.  Now  take  a  stick  of  sodium  hydroxide 
1  in.  long,  "  pure  by  alcohol  "  (note  1),  wash  the  carbonate  from  the 
surface  (note  2),  and  dissolve  in  25  or  30  cc.  of  distilled  water.  Next 
mix  the  two  solutions  in  a  300-cc.  casserole,  and  stir  well,  testing  also  to 
see  if  there  is  an  excess  of  the  alkali,  and  adding  more  if  necessary. 
(Why  have  an  excess?)  After  the  silver  oxide  has  settled  down  almost 
completely,  carefully  pour  off  the  liquid  without  wasting  any  more  of  the 
precipitate  than  can  be  avoided.  Wash  the  silver  oxide  four  times  by 
decantation;  that  is,  add  about  200  cc.  of  distilled  water,  stir  thoroughly 
and  allow  to  settle,  and  then  pour  off  the  liquid  as  at  first,  repeating 
the  process  four  times.  This  process  should  remove  all  the  sodium 


THE  COMPOSITION  OF  SILVER  OXIDE  23 

nitrate  and  the  excess  of  alkali.  Drain  off  the  water,  and  then  dry  the 
product  on  a  steam  bath.  To  remove  the  last  traces  of  water,  powder 
the  oxide  in  a  small  evaporating  dish,  and  then  heat  in  the  air  bath  for 
one  hour  at  a  temperature  of  148°  C.  Great  care  should  be  taken  to 
maintain  the  temperature  and  not  allow  it  to  rise  higher  (note  3).  If 
at  any  time  it  is  seen  to  be  going  too  high,  it  may  be  checked  instantly 
by  removing  the  cover  of  the  air  bath. 

As  soon  as  the  heating  is  completed,  transfer  the  silver  oxide  to  a 
weighing  bottle,  and  as  a  double  precaution  against  moisture  keep  this  in 
a  desiccator. 

When  the  silver  oxide  is  ready,  clean  and  ignite  a  No.  0  porcelain 
crucible.  Allow  this  to  cool  somewhat  on  the  triangle  (note  4),  remove 
it  to  a  desiccator,  and  when  perfectly  cold  (note  5),  weigh  accurately 
to  the  fourth  decimal  place.  After  recording  the  weight  of  the  crucible, 
place  another  1-gm.  weight  on  the  balance  pan,  and  then  place  in  the 
crucible  enough  of  the  silver  oxide  to  again  restore  equilibrium.  Great 
care  must  be  taken  not  to  get  a  particle  of  the  oxide  on  the  balance  pan, 
or  the  weight  of  the  sample  will  be  in  error  by  just  this  amount.  In 
getting  the  final  adjustment  of  the  equilibrium  very  exactly,  a  steel 
spatula  must  be  used,  as  this  makes  it  possible  to  transfer  very  small 
amounts  of  the  oxide. 

Now  remove  the  crucible  to  a  triangle  resting  on  a  ring  stand,  cover, 
and  heat,  at  first  gently,  with  the  flame  some  distance  below  the  crucible. 
Finally,  when  most  of  the  oxide  is  decomposed,  heat  to  faint  redness  for  a 
moment.  Allow  to  cool  as  mentioned  above,  and  when  perfectly  cold, 
weigh. 

From  the  data  obtained  in  this  experiment  calculate  the  percentage 
of  silver  in  silver  oxide.  The  exact  value  is  93.09  per  cent. 

Save  the  pure  silver  formed  in  this  experiment  for  use  in  the  next 
experiment. 

NOTES. — (1)  "  Pure  by  alcohol  "  means  that  the  sodium  hydroxide  has  been  dis- 
solved in  alcohol,  and  thus  separated  from  such  impurities  as  sodium  chloride  which 
do  not  thus  dissolve.  If  chloride  were  present,  the  silver  oxide  would  be  mixed  with 
silver  chloride.  (How  would  you  test  for  chloride  in  the  NaOH?) 

(2)  Because  of  the  action  of  the  carbon  dioxide  in  the  air,  sodium  hydroxide 
always  contains  some  carbonate,  especially  on  the  surface  of  the  sticks.     This  may  be 
dissolved  off  and  thrown  away. 

(3)  Silver  oxide  is  decomposed  completely  into  silver  and  oxygen  at  250°  C.     It 
probably  begins  to  decompose  at  200°  C.     At  145°-150°,  however,  there  seems  to  be 
little  or  no  decomposition,  and  this  temperature  is  necessary  to  get  rid  of  all  the  water. 

(4)  If  a  red-hot  crucible  is  put  immediately  into  a  desiccator  the  latter  is  heated 
so  much  that  a  long  time  is  required  for  the  temperature  to  become  normal  again. 
It  is  best  to  allow  the  crucible  to  cool  so  that  it  can  barely  be  touched  without  dis- 
comfort. 


24 


THE  LAWS  OF  COMBINATION 


(5)  If  an  object  is  weighed  while  still  warm,  an  upward  current  of  air  will  be 
developed  around  the  balance  pan,  and  the  object  will  appear  to  weigh  less  than  it 
really  does. 

Exp.    13.  Composition  of  Silver  Chloride. 

Apparatus. — A  Gooch  perforated  crucible,  a  suction  flask  to  go  with 
it,  a  water  suction  pump,  and  a  wash  bottle. 

The  Gooch  is  a  slender  porcelain  crucible  with  a  perforated  sieve-like 
bottom  upon  which  a  mat  of  asbestos  is  formed.  This,  when  held  in 
place  by  a  porcelain  disk  or  a  layer  of  glass  beads,  forms  the  filtering 
medium.  When  properly  constructed,  this  mat  filters  as  well  as  the  best 
paper,  and  has  two  additional  advantages :  it  does  not  break  when  suc- 
tion is  employed,  and  it  can  be  directly  weighed.  If  a  paper  is  used 
it  must  be  gotten  rid  of  by  burning  before  the  substance  upon  it  can 
be  weighed,  and  in  many  cases  the  burning  of  the  paper  causes  serious 
change  in  this  substance. 

For  the  preparation  of  the  Gooch  filter,  consult  the  instructor.  The 
general  procedure  is,  however,  as  follows:  Fasten  the  crucible  in  place 
by  means  of  the  rubber  stopper  or  a  piece  of  rubber  tubing  as  shown  in 
Fig.  13,  apply  suction,  and  float  in  enough  asbestos  to  make  a  mat 
about  1/32  inch  thick  (note  1).  Place  upon  this  mat  the  disk  or  beads, 
and  then  wash  with  distilled  water.  Break  the  suction  by  pulling  off 


FIG.  13. 


FIG.  14. 


the  rubber  tube  attached  at  A  (not  by  turning  off  the  pump),  remove  the 
crucible,  wipe  with  a  clean  towel,  and  then  dry  completely  by  setting 
in  a  common  crucible  and  heating  gently  on  a  triangle.  The  outside 
crucible  may  finally  be  heated  nearly  to  redness.  Cool  in  a  desiccator 
for  the  usual  time,  and  then  weigh. 

The  wash  bottle  is  fitted  up  according  to  Fig.  14.     The  stopper  is 


COMPOSITION  OF  SILVER  CHLORIDE  25 

of  rubber,  not  cork.  In  bending  the  tubes  use  a  flat  flame,  not  a  Bunsen 
flame.  The  bends  must  be  perfectly  smooth  and  round,  and  all  ends 
must  be  fire-polished.  The  neck  of  the  flask  is  wrapped  with  candle 
wicking  so  that  it  may  be  handled  while  containing  hot  water. 

After  fitting  up  the  apparatus  as  directed,  have  it  inspected  before 
use. 

Procedure. — First  prepare  and  weigh  the  Gooch  crucible  as  directed 
above,  then  return  it  to  the  desiccator  and  allow  it  to  remain  there  until 
needed. 

Take  the  pure  silver  formed  in  Exp.  13  and  already  weighed,  carefully 
loosen  from  the  crucible  by  means  of  a  steel  spatula,  and  transfer  to  a 
300-cc.  beaker.  Be  very  sure  that  this  removal  is  complete  and  that  no 
particles  are  lost.  Pour  over  the  silver  in  the  beaker  10  cc.  of  1  :  1 
nitric  acid,  and  immediately  cover.  Also  rinse  the  crucible  with  a  little 
of  this  acid,  and  add  this  to  the  main  portion.  Warm  the  beaker,  if 
necessary,  to  hasten  solution.  When  every  trace  of  the  silver  is  dis- 
solved, wash  down  the  cover  glass  and  the  sides  of  the  beaker  with  dis- 
tilled water,  and  then  dilute  the  solution  to  about  150  cc.  (Do  not 
measure.) 

Now  heat  the  solution  nearly  to  boiling,  and  then  add  slowly,  with 
constant  stirring,  5  cc.  of  6  N  hydrochloric  acid.  Continue  the  stirring 
and  heating  until  the  precipitate  has  thoroughly  coagulated  and  the 
liquid  above  it  is  almost  clear.  It  is  then  possible  to  tell  by  adding  a 
little  more  acid  whether  precipitation  is  complete.  If,  when  the  acid  is 
added,  a  turbidity  is  produced,  more  acid  must  be  added  until  the  pre- 
cipitation is  complete. 

When  precipitation  is  complete  put  the  weighed  Gooch  crucible  in 
place  on  the  suction  flask,  start  the  pump,  and  then  filter  the  solution 
as  rapidly  as  it  can  be  drawn  through.  Finally  wash  the  chloride  into 
the  crucible,  cleaning  out  the  last  traces  adhering  to  the  glass  by  means  of 
a  rubber-tipped  rod  (a  "  policeman")-  Since  no  solids  are  present  in 
the  solution,  it  is  not  necessary  to  wash  the  chloride  any  more  than  will 
be  done  by  getting  it  all  into  the  crucible. 

When  the  pump  has  removed  as  much  of  the  water  as  possible  from 
the  chloride,  remove  the  crucible,  and  dry  as  noted  above,  only  do  not 
heat  so  strongly  (note  2).  The  safest  way  is  to  put  the  crucible  in  the 
air  bath  and  heat  to  about  200°  C.  for  half  an  hour.  After  being  thus 
heated  and  cooled,  the  crucible  and  contents  are  weighed,  and  then  the 
heating  is  repeated;  if  on  second  heating  there  is  no  further  loss  in 
weight,  the  process  is  stopped.  This  is  called  "  heating  to  constant 
weight." 

Having  determined  the  weight  of  the  silver  chloride,  calculate  from 


26  THE  LAWS  OF  COMBINATION 

this  and  the  weight  of  silver  taken  the  percentage  composition  of  silver 
chloride.  The  accepted  values  are:  silver  75.26  per  cent,  chlorine  24.74 
per  cent.  Your  values  should  not  differ  from  these  by  more  than 
0.2  per  cent. 

Place  the  silver  chloride  in  the  bottle  marked  "  silver  residues." 

NOTES. — (1)  The  asbestos  used  in  a  Gooch  filter  is  prepared  especially  for  the 
purpose  by  boiling  up  with  dilute  hydrochloric  acid,  and  then  washing  to  remove  any 
soluble  matter.  It  is  then  stirred  up  with  a  large  amount  of  water,  and  both  the 
finest  and  the  coarsest  fibers  are  rejected. 

(2)  Silver  chloride  melts  at  460°  C.;  if  heated  to  a  much  higher  temperature,  it 
volatilizes.  It  is  usually  considered  allowable  to  heat  until  it  just  begins  to  melt 
around  the  edges. 

Exp.  14.  The  Multiple  Proportions  of  Chlorine  in  the  Chlorides  of 
of  Mercury. 

Procedure  A. — First  prepare  and  weigh  a  Gooch  crucible.  Next, 
clean,  dry  and  weigh  a  7-cm.  porcelain  evaporating  dish.  When  the 
correct  weight  has  been  added  to  balance  the  dish,  add  another  1-gm. 
weight,  and  then  by  means  of  a  steel  spatula  carefully  place  in  the  dish 
enough  mercurous  chloride  to  restore  equilibrium.  Record  only  the 
weight  of  the  sample  thus  weighed. 

Now  treat  the  chloride  with  a  solution  containing  0.5  gm.  of  sodium 
hydroxide  (free  from  chloride)  in  25  cc.  of  water,  and  heat  on  the  steam 
bath  for  fifteen  minutes  (note  1)  with  frequent  stirring.  (Do  not  with- 
draw any  of  the  solution  when  removing  the  rod.)  At  the  end  of  this 
time  dilute  the  solution  somewhat  and  then  filter  and  wash  the  pre- 
cipitate until  free  from  chloride. 

To  the  alkaline  filtrate  containing  the  Chloride  now  add  dilute  nitric 
acid  until,  after  thorough  stirring,  a  minute  piece  of  litmus  paper  pre- 
viously placed  in  the  liquid  shows  a  distinct  acid  reaction.  (Why  is 
this  done?) 

Dilute  the  solution  (if  necessary)  to  about  150  cc.,  heat  nearly  to 
boiling  and  then  precipitate  the  chloride  radical  as  follows:  Add,  with 
stirring,  25  cc.  of  N/5  silver  nitrate  solution,  and  then  continue  the  beat- 
ing and  stirring  until  the  solution  clears.  Test  then  for  complete  pre- 
cipitation by  use  of  a  few  more  drops  of  the  silver  nitrate.  Be  sure  that 
the  precipitation  is  complete,  but  do  not  add  an  excess  of  the  reagent. 

Proceed  now  with  the  filtration,  as  directed  under  the  last  experiment, 
but  remember  that  in  this  case  the  solution  contains  salts  (what  salts?) 
which  must  be  washed  out.  Five  or  six  washings  with  hot  water  will 
not  be  too  much.  The  drying  and  weighing  are  also  conducted  as  under 
the  last  experiment. 

After  the  weight  of  the  silver  chloride  is  obtained  find  the  weight 


VOLUMES  FROM  THE  ANALYSIS  OF  AMMONIA  27 

of  chlorine  originally  contained  in  1  gm.  of  mercurous  chloride;  and 
finally  calculate  by  proportion  the  weight  of  chlorine  which  would  be 
combined  with  I  gm.  of  mercury  in  this  compound. 

Procedure  B. — Weigh  out  a  1-gm.  sample  of  mercuric  chloride, 
proceeding  as  above.  Dissolve  in  20  cc.  of  hot  water,  add  a  solution 
containing  1  gm.  of  sodium  hydroxide,  stir,  and  heat  on  a  steam  bath 
until  the  precipitate  has  thoroughly  settled  (note  2).  From  this  point 
proceed  as  directed  in  Procedure  A,  with  the  following  exception: 
Use  40  cc.  of  N/5  silver  nitrate  solution  in  precipitating  the  chloride 
radical,  instead  of  25  cc.  The  Gooch  crucible  may  be  used  with  the 
former  precipitate  in  it. 

After  the  weight  of  the  silver  chloride  is  found,  calculate  as  above  the 
weight  of  chlorine  combined  with  I  gm.  of  mercury  in  mercuric  chloride. 

Having  analyzed  the  two  chlorides  of  mercury,  note  the  relationship 
between  the  two  weights  of  chlorine,  each  combined  with  one  and  the 
same  weight  (1  gm.)  of  mercury.  Calculate  also  the  equivalent  weight 
of  mercury  in  the  two  cases. 

Please  put  the  silver  chloride  in  the  silver  residues  bottle. 

NOTES. — (1)  Although  the  mercurous  chloride  is  nearly  insoluble  in  water,  it 
reacts  with  sodium  hydroxide  according  to  the  equation 

2NaOH+2HgCl  -»  2NaCl+Hg,O+H2O 

and  if  time  enough  is  allowed,  the  reaction  is  nearly  complete.  Heat  hastens  the 
process.  The  chloride  radical,  which  is  to  be  determined,  is  thus  transferred  from 
the  mercury  to  the  sodium,  and  is,  of  course,  found  in  the  filtrate  with  the  excess  of 
alkali.  It  is  necessary  to  get  it  into  this  form  before  it  can  be  precipitated  with  silver 
nitrate,  as  this  reagent  does  not  react  with  mercurous  chloride. 

(2)  If  the  mercuric  chloride  were  finely  powdered  the  reaction  would  slowly 
proceed  to  completion  upon  simply  treating  the  solid  directly  with  the  alkali;  but  it 
is  always  best  to  carry  out  such  an  operation  in  solution  where  the  reaction  is  practi- 
cally instantaneous.  This  reaction  is  represented  by  the  equation 

HgCl2+2NaOH  — >  HgO+2NaCl+H2O 

The  treatment  with  NaOH  is  necessary,  even  in  the  case  of  this  soluble  chloride, 
since  silver  nitrate  does  not  completely  precipitate  the  chlorine  in  presence  of 
mercury. 

Exp.  15.  The  Law  of  Volumes  from  the  Analysis  of  Ammonia. 

Ammonia  gas,  generated  by  heating  strong  ammonium  hydroxide 
and  dried  by  passing  over  soda-lime,  is  led  over  hot  copper  oxide,  where 
the  hydrogen  is  removed  to  form  water.  The  nitrogen  passes  on  and  is 
measured.  Data  are  also  obtained  as  to  the  volume  of  hydrogen  orig- 
inally combined  with  the  nitrogen  and  of  the  ammonia  decomposed. 


28 


THE  LAWS  OF  COMBINATION 


Apparatus. — A  250-cc.  Erlenmeyer  flask,  a  6-inch  U-tube,  and  an 
8-inch  combustion  tube  of  thin,  hard  glass  with  tapering  ends,  as  shown 
in  Fig.  15.  Any  bottle  will  do  for  collecting  the  gas,  e.g.,  one  of  the 
600-cc.  glass-stoppered  bottles  of  the  regular  outfit.  All  connections 


E 


FIG.  15. 

must  be  perfectly  secure.     Use  rubber  stoppers,  and  test  the  apparatus 
for  leaks  before  use. 

Procedure. — Fill  the  combustion  tube  with  copper  oxide  (the  wire 
form)  and  fasten  in  place  by  means  of  a  wad  of  long  fiber  asbestos  placed 
near  each  end.  To  make  sure  that  this  oxide  contains  no  water,  connect 
the  tube  to  a  pump  and  draw  through  it  a  slow  current  of  dry  air  while 
heating  moderately  with  a  burner.  Allow  the  tube. to  cool,  and  weigh 
when  perfectly  cold. 

While  the  tube  is  cooling  fill  the  U-tube  with  granulated  soda-lime, 
and  close  with  rubber  stoppers  (note  1).  Also  place  in  the  flask  about 
100  cc.  of  the  strongest  ammonium  hydroxide  solution. 

Finally,  when  the  combustion  tube  has  been  weighed,  connect  the 
parts  of  the  apparatus  as  seen  in  the  sketch.  Do  not  put  the  delivery 
tube  under  the  bottle,  and  do  not  forget  the  pinch  cocks. 

Now  warm  the  flask  gently  (note  2)  and  watch  the  bubbles  as  they 
issue  from  the  delivery  tube.  Air  is  insoluble  (nearly)  in  water,  but 
ammonia  is  very  soluble.  Therefore,  when  all  the  air  is  expelled  from 
the  apparatus  the  bubbles  will  cease  to  rise  through  the  water.  When 
this  occurs  place  the  delivery  tube  under  the  bottle  and  then  imme- 
diately begin  heating  the  combustion  tube.  The  heating  should  be 
gentle  at  first,  the  flame  being  waved  back  and  forth,  and  great  care 
should  be  taken  not  to  burn  the  rubber  connectors.  Later,  heat  strongly 
until  the  oxide  decomposes  the  ammonia,  and  nitrogen  is  evolved  (note  3) 
Do  not  allow  the  water  to  condense  at  the  end  of  the  tube.  When  the 
bottle  is  about  one-third  full,  or  when  the  copper  oxide  is  nearly  all 
reduced,  stop  heating  the  tube,  but  allow  the  ammonia  to  pass  until 
all  the  nitrogen  is  swept  over.  Finally  stop  warming  the  ammonia  flask, 
carefully  watch  the  delivery  tube  until  the  water  begins  to  recede 
(Why  does  it  recede?)  and  then  quickly  put  the  clip  E  in  place.  Allow 


VOLUMES  FROM  THE  ANALYSIS  OF  AMMONIA  29 

the  tube  to  cool  without  disconnecting  (note  4).  When  cold,  disconnect, 
draw  out  the  ammonia  with  a  pump  (Why?),  and  then  weigh. 

Since  a  good  deal  of  ammonia  has  been  run  into  the  pneumatic 
trough,  the  nitrogen  collected  will  contain  some  ammonia,  and  this 
must  be  removed  before  the  nitrogen  can  be  measured.  To  accomplish 
this,  stopper  the  bottle  securely,  and  transfer  to  another  trough  con- 
taining clean  water,  setting  over  one  of  the  holes  in  the  shelf.  To  hasten 
the  removal  of  the  ammonia,  attach  a  rubber  tube  to  the  water  faucet 
and  run  a  stream  of  water  up  into  the  bottle  from  below,  first  driving 
the  air  from  the  tube,  of  course. 

Finally  measure  the  nitrogen  in  the  usual  way,  and  calculate  the 
volume  down  to  standard  conditions,  not  forgetting  to  allow  for  water 
vapor.  Calculate  also  the  weight  of  the  nitrogen  from  its  volume.  . 

The  loss  in  weight  suffered  by  the  copper  oxide  represents  the  oxygen 
removed  from  it.  This  oxygen  has  united  with  the  hydrogen  of  the 
ammonia  to  form  water.  Knowing  the  composition  of  water,  you  can 
immediately  calculate  the  weight  of  this  hydrogen,  and  from  its  weight 
you  can  then  calculate  its  volume  under  standard  conditions.  Remem- 
ber that  this  is  the  hydrogen  which  was  originally  combined  with  the 
nitrogen  you  have  collected. 

Adding  together  the  weights  of  the  nitrogen  and  the  hydrogen  will 
give  the  weight  of  the  ammonia  decomposed.  From  this  weight  the 
volume  may  be  calculated  as  in  the  other  cases. 

Having  thus  found  the  volumes  of  the  nitrogen  and  hydrogen  and 
of  the  ammonia  from  which  they  came,  you  will  at  once  notice  the 
extremely  simple  and  exact  relationship.  State  this  relationship. 

NOTES. — (1)  Soda  lime  is  used  to  dry  the  ammonia,  since  calcium  chloride  or 
sulphuric  acid  would  combine  with  it.  "  Quicklime  "  (calcium  oxide)  might  be  used, 
but  is  not  so  good. 

(2)  Use  a  flame  about  half  an  inch  high.     Great  care  must  be  taken  that  the 
warming  is  not  interrupted  by  draughts  of  air  or  otherwise.     If  this  happens,  water 
will  be  sucked  back  into  the  combustion  tube. 

(3)  Copper  oxide  reacts  with  ammonia  thus: 

3CuO+2NH3  ->  3Cu+3H2O+N2 

(4)  If  the  tube  is  opened  while  still  hot,  air  will  enter  and  oxidize  the  copper,  thus 
changing  its  weight. 


CHAPTER  VI 


ATOMIC  WEIGHTS 

Exp.  16.  Specific  Heat  and  Atomic  Weight  of  Tin. 

The  specific  heat  of  tin  is  determined  by  the  so-called  "  method  of 
mixtures."  That  is,  a  known  weight  of  tin  is  heated  to  a  known  tem- 
perature and  then  dropped  into  a  known  weight  of  water,  also  at  a 
known  temperature.  From  the  rise  in  the  temperature  -of  the  water  we 
calculate  the  heat  given  off  by  1  gm.  of  tin  in  cooling  1°  C. — the  specific 
heat. 

Apparatus. — The  calorimeter  arrangement  of  Exp.  9,  with  a  200-cc. 
beaker  instead  of  the  400-cc.  beaker  there  recommended;  a  delicate 
thermometer  graduated  in  0.1°.  For  heating  the  tin,  a  large  test-tube 
(25X180  mm.)  is  suspended  in  an  Erlenmeyer  flask 
as  seen  in  Fig.  16,  the  latter  serving  as  a  steam 
jacket.  To  prevent  the  test-tube  from  dropping 
down  too  far  in  the  flask,  a  filter  paper  is  folded 
about  it  and  tied  in  place  as  indicated.  This  also 
serves  to  catch  any  drops  of  water  which  might  run 
off  into  the  calorimeter  when  the  tin  is  poured  out. 
For  taking  the  temperature  of  the  tin  a  common, 
but  accurate,  thermometer  will  be  needed. 

Procedure. — Weigh  the  calorimeter  beaker,  put 
into  it  100  cc.  of  water,  and  weigh  again.  Now 
place  this  in  the  graniteware  cup  with  proper  insu- 
lation, and  then  suspend  within  it  the  tenth-degree 
thermometer.  (To  prevent  breakage  it  will  be  well 
to  suspend  the  thermometer  by  means  of  a  strong 
cord,  not  by  means  of  a  clamp,  which  is  likely  to  slip.)  After  stirring 
the  water  for  a  few  minutes  by  means  of  the  thermometer,  the  bulb 
of  which  should  be  entirely  immersed,  take  the  temperature  as  accu- 
rately as  possible.  This  should  be  done  by  means  of  a  lens,  and  the 
thermometer  should  be  tapped  gently  with  a  pencil  just  before  reading. 
Avoid  parallax. 

Now  weigh  out  on  the  laboratory  balance  about  100  gm.  of  granulated 
tin  (of  about  30-mesh  size)  and  place  it  in  the  test-tube  for  heating. 

30 


FIG.  16. 


SPECIFIC  HEAT  AND  ATOMIC  WEIGHT  OF  TIN  31 

Put  about  1  inch  of  water  in  the  Erlenmeyer,  embed  the  thermometer  in 
the  tin,  plug  the  mouth  of  the  test-tube  loosely  with  cotton,  and  then 
suspend  it  in  the  flask  as  seen  in  the  sketch.  Finally  heat  the  water  to 
rapid  boiling. 

When  the  temperature  of  the  tin  has  reached  its  maximum  (about 
99°),  read  the  thermometer  in  the  calorimeter  once  more,  withdraw  the 
thermometer  from  the  test-tube,  remove  the  cotton,  and  instantly  pour 
the  tin  into  the  calorimeter,  tapping  the  tube  to  insure  complete  removal. 
Now  stir  with  the  thermometer  until  the  temperature  of  the  mixture  is 
uniform  (perhaps  30  sec.),  taking  care  to  stir  the  tin  as  well  as  the  water. 
Finally  read  the  thermometer  accurately,  as  at  first. 

You  now  have  all  the  necessary  data.  Calculate  first  the  water 
equivalent  of  the  calorimeter  and  add  this  to  the  weight  of  the  water, 
calculate  next  the  change  in  the  temperature  of  the  water  and.  calorim- 
eter, and  then  the  number  of  calories  of  heat  absorbed  by  them.  This 
heat  all  came  from  the  tin  in  cooling  through  a  considerable  number  of 
degrees.  Find  what  has  been  the  total  change  in  the  temperature  of  the 
tin,  and  then  calculate  the  number  of  calories  it  gave  off  in  changing  one 
degree.  Finally  calculate  the  heat  liberated  by  1  gm.  of  tin  in  cooling 
one  degree.  This  is  its  specific  heat. 

Now  proceed  with  the  calculation  of  the  atomic  weight  of  tin  as 
follows : 

(1)  Apply  Dulong  and  Petit's  law  for  the    determination  of  the 
approximate  atomic  weight. 

(2)  Calculate  the  equivalent  weight  of  tin  from  the  following  data: 
9.8137  gm.  of  tin  were  oxidized  to  stannic  oxide.     The  weight  of  the 
oxide  was  12.4598  gm. 

(3)  Compare  the  equivalent  weight  with  the  approximate  atomic 
weight,  and  thus  determine  what  multiple  of  the  equivalent  weight  to 
use  for  the  exact  atomic  weight.     Calculate  then  the  exact  atomic 
weight. 


CHAPTER  VII 


VALENCE 

Exp.  17.  Valence  of  Sodium,  Magnesium,  and  Aluminum. 

This  group  of  experiments  is  intended  to  direct  attention  to  the 
simple  idea  that  valence  is  measured  by  the  combining  or  displacing 
capacity  which  an  element  has  towards  hydrogen.  Weights  of  metals 
are  taken  which  are  exactly  proportional  to  their  atomic  weights.  This 
insures  the  presence  of  the  same  number  of  atoms  of  each  metal.  We 
allow  these  metals  to  react  with  an  excess  of  water  or  hydrochloric  acid 
until  entirely  consumed.  The  volume  of  hydrogen  obtained  in  each 
case  is  a  measure  of  the  number  of  atoms  displaced,  and  therefore  of  the 
valence.  We  also  note  that  the  relation  here  found  is  the  same  as  the 
relation  between  the  atomic  weights  and  the  equivalent  weights. 

Apparatus. — As  seen  in  Fig.  17.      The  flask  should  not  be  larger 

than  200  cc.,  but  the  neck 
should  be  large  enough  to 
take  a  No.  4  two-hole  rub- 
ber stopper. 

The  tap  funnel  holds 
60-75  cc.  and  has  a  stem 
small  enoughto  pass  through 
the  stopper  easily  when  wet. 
Be  sure  that  the  stop-cock 
does  not  leak,  lubricating  if 
necessary. 

The  outlet  tube  must 
not  project  below  the  stopper,  or  hydrogen  will  be  caught  beside  it  and 
cannot  be  removed. 

The  piece  of  rubber  tubing  at  the  end  of  the  funnel  tube  is  intended 
to  prevent  hydrogen  going  up  the  tube  and  being  trapped  there.  It 
should  not  kink  so  as  to  close  the  tube. 

Sodium,  one  of  the  metals  used,  must  be  protected  while  being 
weighed.  This  is  done  by  enclosing  it  in  a  gelatin  capsule.  Use  size 
No.  1 ;  a  larger  size  would  permit  the  use  of  too  large  a  piece  of  sodium, 
and  make  the  experiment  dangerous. 

32 


FIG.  17. 


VALENCE  OF  SODIUM,  MAGNESIUM,  AND  ALUMINUM          33 

Procedure  A.  Sodium. — First  fill  the  pneumatic  trough  and  place 
the  bottle  for  the  collection  of  hydrogen.  Next  weigh  the  gelatin  cap- 
sule accurately,  place  within  it  a  single,  cleanly  cut  piece  of  sodium  as 
large  as  possible  (instructor),  and  then  cover  and  weigh  as  before. 
When  ready,  place  the  loaded  capsule  under  the  mouth  of  the  bottle, 
and  allow  it  to  rise  to  the  surface  of  the  water  inside.  Within  about  ten 
minutes  the  capsule  will  dissolve  and  allow  the  sodium  to  react  with  the 
water.  When  the  reaction  begins  it  is  well  to  throw  a  towel  over  the 
bottle,  so  that  in  case  of  a  possible  small  explosion  no  harm  can  be  done. 
When  the  reaction  is  over,  adjust  levels,  stopper  the  bottle,  and  then 
proceed  in  the  usual  way  to  measure  the  hydrogen.  Finally  calculate 
its  volume  down  to  standard  conditions,  not  forgetting  to  allow  for 
water  vapor. 

Also  calculate  from  your  data  the  equivalent  weight  of  sodium. 

Procedure  B.  Magnesium. — First  calculate  what  part  of  an 
atomic  weight  of  sodium  you  used  in  Procedure  A,  and  then  calculate 
what  weight  of  magnesium  will  represent  the  same  fraction  of  its  atomic 
weight.  To  get  exactly  this  weight  of  the  metal,  carefully  clean  a  piece 
of  the  wire  of  exact  known  length  (10  cm.)  with  emery  cloth,  and  then 
weigh.  It  is  then  only  the  matter  of  a  moment  to  calculate  the  length 
necessary  to  give  the  proper  weight.  The  wire  should  be  straight,  of 
course,  when  measured,  and  the  ends  should  be  cut  off  square.  The 
cutting  can  best  be  done  by  rolling  under  the  edge  of  a  knife.  If  this 
work  is  carefully  done  no  after  weighing  is  necessary. 

Fill  the  flask  A  full  of  water,  drop  in  the  magnesium  wire,  and  push 
the  stopper  with  the  connecting  tubes  in  place.  Now  run  water  through 
the  apparatus  from  the  tap  funnel  until  all  the  air  is  removed  from  the 
neck  of  the  flask  and  from  the  delivery  tube,  but  close  the  cock  while 
there  is  still  a  very  little  water  above  it;  that  is,  do  not  allow  air  to 
get  in. 

Now  put  the  bottle  B  in  place,  and  run  perhaps  30  cc.  of  concen- 
trated hydrochloric  acid  into  the  flask,  taking  care  not  to  let  any  air  in. 
After  the  magnesium  is  completely  dissolved,  drive  over  the  last  of  the 
hydrogen  by  running  water  through  the  apparatus  as  at  first.  Finally 
measure  the  mixture  of  hydrogen  and  water  vapor,  and  calculate  the 
volume  of  the  latter  down  to  0°  and  760  mm. 

Calculate  also  the  equivalent  weight  of  magnesium. 

How  many  equivalent  weights  does  the  atomic  weight  of  magnesium 
contain? 

Assuming  that  the  volume  of  hydrogen  obtained  in  the  case  of  sodium 
represents  "  one  atom,"  how  many  atoms  of  hydrogen  does  a  like  atomic 
proportion  of  magnesium  displace? 


34  VALENCE 

Procedure  C.  Aluminum. — Using  the  same  apparatus  and  pro- 
ceeding in  the  same  way  as  in  B,  obtain  the  same  data  for  aluminum. 
It  may  be  necessary  to  heat  the  flask  to  start  the  reaction.  The  water 
should  not  be  boiled,  however. 

How  many  equivalent  weights  does  the  atomic  weight  of  aluminum 
contain? 

How  many  "  atoms  "  of  hydrogen  do  you  get,  as  compared  with  the 
case  of  sodium? 

When  all  the  data  for  the  three  metals  have  been  obtained,  arrange 
in  tabular  form  as  follows: 

Na.      Mg.      Al. 

(1)  Weight  of  metal  used 

(2)  Fraction  of  an  atomic  weight 

(3)  Volume  of  hydrogen  at  0°  and  760  mm 

(4)  Relative  volumes  (vol.  with  sodium  =  1) 

(5)  Valence,  from  relative  volumes 

(6)  Equivalent  weights 

(7)  Atomic  weight 

(8)  Number  of  equivalents  in  at.  wt 

(9)  Valence,  from  (8) 

Exp.  18.  Oxidation  and  Reduction  Valence. 

The  following  experiments  are  intended  to  show  how  we  determine 
the  active  valence  of  oxidizing  and  reducing  agents.  We  shall  make  up 
solutions  containing  1/10  mole  *  of  each  compound  per  liter.  We  shall 
then  have  present  in  each  case  the  same  number  of  molecules  per  cubic 
centimeter.  If,  then  1  cc.  of  solution  A  reacts  with  1  cc.  of  solution  B,  we 
know  that  1  molecule  of  substance  A  reacts  with  1  molecule  of  substance 
B;  and,  that,  therefore,  the  active  valence  of  A  and  B  are  equal,  one 
as  an  oxidizer,  the  other  as  a  reducer.  If  1  cc.  of  solution  A  reacts  with 
5  cc.  of  solution  B,  we  know  that  1  molecule  of  A  reacts  with  5  molecules 
of  B;  and  this  shows  that  the  valence  of  A  is  five  times  that  of  B. 

We  shall  use  as  our  standard  a  solution  of  iodine.  The  active  valence 
of  iodine  as  an  oxidizing  agent  is  1.  This  is  shown  by  the  fact  that  the 
neutral,  uncombined  iodine,  I,  as  seen  in  its  brown  or  violet  solutions 
changes  to  the  colorless  ion  I~  when  it  goes  into  combination,  to.  form, 
for  example,  HI  or  KI.  The  fact  that  the  charge  becomes  negative 
indicates  that  iodine  is  an  oxidizing  agent. 

The  standard  iodine  solution  will  be  furnished.  Directions  for  pre- 
paring the  other  solutions  are  included. 

Apparatus. — The  ordinary  volumetric  apparatus,  including  one 
burette  with  glass  stopper,  one  burette  with  rubber  tip,  a  set  of  five 
pipettes,  and  one  or  two  graduated  flasks. 

*  1/10  atomic  weight  in  the  case  of  elements. 


OXIDATION  AND  REDUCTION  VALENCE  35 

In  using  the  volumetric  apparatus,  the  following  suggestions  should 
be  followed: 

(1)  The  inside  of  the  glass  should  be  so  clean  that  drops  will  not  form 
on  the  sides  when  the  solutions  are  run  out.     (Why?) 

(2)  The  stop-cock  of  the  burette  should  be  slightly  lubricated  to 
prevent  setting  and  leakage. 

(3)  The  Mohr  burette  should  be  provided  with  a  glass  tip,  such  as  is 
used  on  a  wash  bottle,  and  with  a  pinch  clamp. 

(4)  When  in  use,  burettes  should  be  clamped  so  as  to  stand  perfectly 
plumb.     Nothing  more  certainly  distinguishes  the  slovenly  worker  than  a 
carelessly  placed  burette. 

(5)  If  a  standard  solution  is  to  be  placed  in  a  wet  burette,  the  latter 
must  first  be  rinsed  with  a  little  of  the  solution.     Otherwise  the  solution 
would  be  diluted,  or  contaminated  with  some  other  solution.     Do  not 
forget  the  part  below  the  stop-cock. 

(6)  Before  beginning  a  titration,  remove  any  bubbles  of  air  from  the 
tip  of  the  burette. 

(7)  Avoid  parallax  in  reading  a  burette. 

(8)  Do  not  put  oxidizing  agents  (e.g.,  KMnCU)   in  rubber-tipped 
burettes. 

(9)  Pipettes  are  calibrated  to  deliver  the  amount  marked  upon  them 
without   rinsing.     After   allowing   a   pipette   to   drain  about    10   sec., 
remove  the  drop  at  the  tip  by  closing  the  top  with  the  finger  and  then 
warming  the  bulb  with  the  hand.     Do  not  blow  through  a  pipette. 

Procedure  A.  Reducing  Valence  of  Sodium  Thiosulphate. — In 
making  up  the  solution  of  thiosulphate  two  students  may  work  together,* 
preparing  500  cc. 

Proceed  as  follows:  Calculate  the  molar  weight  of  the  hydrated  salt 
Na2S2O3-5H2O,  and  from  this  calculate  the  amount  necessary  for 
500  cc.  M/10.  Select  clear  crystals  of  thiosulphate — those  which  have 
not  lost  any  water  of  hydration — crush  in  a  mortar,  and  then  weigh  out 
(as  above)  the  calculated  amount.  Dissolve  in  a  small  amount  of  water 
and  make  up  to  500  cc.  in  a  graduated  flask.  Do  not  forget  to  mix  well 
after  making  up  to  volume. 

Now  fill  one  burette  with  the  iodine  solution  and  the  other 
with  the  thiosulphate,  first  reading  over  all  the  precautions  about  the  use 
of  burettes.  Run  off  into  a  beaker  25  cc.  of  the  iodine  solution  and  drop 
the  thiosulphate  solution  into  this  with  constant  stirring  until  the  color 
of  the  iodine  is  just  discharged.  To  make  sure  of  the  end  point,  titrate 
back  with  a  drop  or  two  of  the  iodine  solution  until  a  faint  color  remains 

*  Students  work  in  groups  only  on  the  preparation  of  the  solutions,  not  in  any  case 
on  the  subsequent  titrations. 


36  VALENCE 

after  stirring,  and  then  discharge  again  with  thiosulphate.  Finally 
take  both  the  readings  accurately,  estimating  to  0.1  of  a  division. 

The  end  point  will  be  much  sharper  if  a  solution  of  starch  paste 
(note  1)  is  added  just  before  the  end  point  is  reached.  Use  about  5  cc. 
of  the  solution  made  up  as  directed.  The  iodine  forms  a  bright  blue 
solution  with  the  starch,  thus  making  it  possible  to  detect  a  much 
smaller  amount  of  iodine  than  could  be  detected  by  means  of  its  own 
color.  The  starch  paste  solution  should  not  be  added,  however,  until 
the  end  point  is  almost  reached  (note  2) .  It  is  always  best  to  use  starch 
in  iodine  titrations,  and  in  some  cases,  e.g.,  with  highly  colored  solutions, 
it  is  absolutely  necessary. 

Now  reread  the  introduction  given  at  the  head  of  this  series  of  pro- 
cedures, and  then  decide  as  to  the  active  valence  of  thiosulphate  as  a 
reducing  agent.  Also  explain  the  above  method. 

NOTES. — (1)  Take  a  piece  of  starch  the  size  of  a  pea,  moisten  with  cold  water, 
and  then  stir  in  very  slowly  about  50  cc.  of  boiling  water.  This  solution  must  be 
used  fresh. 

(2)  If  the  starch  is  added  when  there  is  still  a  large  amount  of  iodine  present,  the 
granules  will  become  so  deeply  impregnated  with  it  that  it  is  not  easily  removed  at 
the  end. 

Procedure  B.  The  Oxidizing  Valence  of  Potassium  Permanganate. 
—Four  students  work  together  on  the  preparation  of  the  solution.  One 
hundred  cc.  will  be  enough.  Obtain  a  graduated  flask  for  the  purpose. 

Calculate  the  weight  of  1/10  mole  of  permanganate,  KMnO4,  and 
take  1/10  of  the  calculated  amount  for  100  cc.  Since  the  amount  to  be 
weighed  here  is  small,  it  will  be  best  to  use  the  accurate  balance,  but  there 
is  no  necessity  for  weighing  beyond  the  second  place.  (What  per  cent 
error  would  0.01  gm.  make  on  1.5  gm.?)  When  the  proper  amount  is 
weighed  out,  dissolve  in  about  20  cc.  of  hot  water,  and  then  transfer  to 
the  graduated  flask.  There  will  be  some  difficulty  in  telling  when  all 
the  solid  is  dissolved,  on  account  of  the  opaqueness  of  the  solution. 
The  particles  may  usually  be  felt,  however,  with  a  rod  (note  1),  and  if 
the  solution  is  carefully  decanted  into  the  graduate  flask,  any  such 
particles  may  then  be  seen.  When  the  salt  is  all  in  solution,  bring  the 
volume  up  to  the  mark  with  water  and  then  mix  carefully. 

To  determine  the  oxidizing  valence  of  permanganate,  measure  out 
carefully  with  a  pipette  10  cc.  of  the  M/10  solution  into  a  beaker.  To 
this  add,  first,  5  cc.  of  6  N  sulphuric  acid,  and  second,  20  cc.  of  a  10  per 
cent  solution  of  potassium  iodide  (note  2).  Dilute  now  to  about  100  cc. 
and  then  titrate  the  iodine  with  thiosulphate  as  directed  under  procedure 
A,  excepting  that  you  leave  the  iodine  solution  where  it  is  in  the  beaker 
and  titrate  it  all  (note  3).  Be  very  careful  not  to  overstep  the  end.  If 


OXIDATION  AND  REDUCTION  VALENCE  37 

you  do,  however,  it  will  be  possible  to  tell  what  excess  of  thiosulphate 
has  been  added  by  titrating  back  with  the  standard  iodine  solution. 

When  through  with  the  titration,  decide  as  to  the  oxidizing  valence 
of  the  permanganate,  and  explain  carefully  the  steps  in  the  above 
method. 

NOTES. — (1)  Do  not  use  a  rubber-tipped  rod.  Permanganate  is  decomposed  by 
rubber  or  any  organic  matter. 

(2)  The  amount  of  potassium  iodide  is  considerably  in  excess  of  the  amount 
required  in  the  reaction,  but  the  excess  is  needed  as  a  solvent  for  the  iodine  which  is 
set  free. 

(3)  Be  very  sure  that  you  know  exactly  what  is  the  purpose  of  all  these  reagents, 
and  also  just  what  is  taking  place  at  all  times.     Do  not  begin  the  experiment  until 
you  do  know  these  things.     Is  it  necessary  that  the  potassium  iodide  and  the  acid  be 
accurately  measured? 

Procedure  C.  The  Reducing  Valence  of  Ferrous  Iron. — As  a  carrier 
for  ferrous  iron,  we  shall  use  the  stable  and  definite  salt,  ferrous  ammo- 
nium sulphate,  FeSO4(NH4)2SO4-6H20,  each  mole  of  which  contains 
one  atomic  weight  of  iron. 

Four  students  working  together  may  prepare  500  cc.  of  the  ferrous 
iron  solution  as  follows:  Calculate  the  weight  of  1/10  mole  of  the  iron- 
carrying  salt  (this  will  give  also  1/10  mole  of  iron)  and  take  half  the 
calculated  amount  for  500  cc.  Dissolve  in  about  200  cc.  of  water  con- 
taining about  5  cc.  of  concentrated  sulphuric  acid  (note  1),  and  then 
make  up  to  volume  and  mix. 

Take  50  cc.  of  the  ferrous  iron  solution  (measured  with  a  pipette), 
add  5  cc.  of  6  N  sulphuric  acid  (note  2)  and  titrate  (note  3)  with  the  per- 
manganate until  a  point  is  reached  where  a  faint  pink  color  remains 
after  thorough  stirring. 

Knowing  the  oxidizing  valence  of  the  permanganate,  you  can  now 
calculate  the  reducing  valence  of  ferrous  iron. 

NOTES. — (1)  If  the  salt  is  dissolved  in  pure  water,  the  iron  in  it  is  slightly  oxidized 
by  the  dissolved  air;  and  since  oxidation  demands  an  increase  in  the  amount  of  the 
negative  radical,  the  only  thing  which  can  happen  is  for  OH  from  the  water  to  unite 
with  the  iron  to  make  up  the  deficiency.  This  forms  a  basic  sulphate  insoluble  in 
water,  and  the  solution  thus  becomes  turbid.  In  presence  of  the  acid  no  basic  sul- 
phate can  be  formed,  and  besides,  permanganate  titrations  must  be  carried  out  in 
acid  solution. 

(2)  KMnO4  has  a  different  valence,  in  neutral  solution. 

(3)  Do  not  use  a  rubber-tipped  burette. 

Procedure  D.    The  Oxidizing  Valence  of  Potassium  Dichromate. — 

Four  students  working  together  prepare  100  cc.  of  M/10  dichromate, 
K2Cr20?.  To  do  this,  calculate  the  molar  weight,  and  then  proceed 
to  weigh  out  1/100  of  this  amount  for  100  cc.  M/10. 


38 


VALENCE 


Place  in  a  beaker  10  cc.  of  10  per  cent  solution  of  potassium  iodide 
and  5  cc.  of  6  N  sulphuric  acid,  add  exactly  5  cc.  of  the  standard  dichro- 
mate,  and,  after  stirring,  titrate  the  iodine  with  thiosulphate.  Because 
of  the  green  color  of  the  Cr+++  it  will  be  necessary  to  use  starch  indicator 
but  this  should  not  be  added  until  the  iodine  is  nearly  all  titrated,  when 
the  red  color  gives  place  to  a  greenish  yellow.  When  the  starch  is  first 
added  the  color  will  probably  be  very  dark  green,  almost  black;  but  as 
the  titration  proceeds  this  color  will  change  to  a  rich  blue;  and  when  the 
end  point  is  reached  the  blue  will  give  place  to  the  light  chromium  green. 

The  relative  volumes  of  dichromate  and  thiosulphate  furnish  the 
means  of  calculating  the  oxidizing  valence  of  the  dichromate.  Having 
made  this  calculation,  explain  also  the  steps  in  the  above  method. 

Exp.  19.  Zinc  as  a  Reducing  Agent. 

Iron  is  commonly  determined  volumetrically  by  reducing  it  to  the 
ferrous  condition  and  then  titrating  with  a  standard  oxidizer,  such  as 
dichromate  or  permanganate.  When  gotten  into  solution  from  its  ores  or 
alloys  iron  is  usually  either  all  ferric  or  a  mixture  of  ferrous  and  ferric. 
To  get  it  all  into  the  ferrous  condition,  one  or  the  other 
of  two  common  reducing  agents  is  used:  these  are 
stannous  chloride  or  zinc.  When  metallic  zinc  acts  as  a 
reducer  the  change  occurring  is, 

2Fe++++Zn  -»  Zn+++2Fe++ 

There  are  several  methods  of  using  zinc  as  a  reducing 
agent,  but  the  one  which  is  probably  the  best  employs 
the  apparatus  known  as  the  "  Jones  reductor  "  (Fig. 
18),  the  preparation  and  use  of  which  will  be  described. 
The  tube  is  50  cm.  long,  including  the  tip  below  the 
stop-cock,  and  2  cm.  in  diameter.  When  in  use  it  is 
supported  by  a  ring  stand.  The  zinc  used  is  30  mesh 
or  slightly  larger,  and  to  avoid  needless  waste  is 
amalgamated.  The  process  of  amalgamation  is  con- 
ducted as  follows:  Dissolve  2  gm.  of  pure  mercury  in 
10  cc.  of  1  :  1  nitric  acid,  dilute  to  200  cc.  in  a  large 
flask,  add  the  zinc  (200  gm.)  and  shake  for  two  minutes. 
Pour  off  the  solution,  wash  the  zinc  thoroughly  and  then 
transfer  to  the  reductor.  To  use  the  reductor  proceed 
as  follows:  Turn  on  the  suction  and  pass  through  the 
tube  100  cc.  of  2  per  cent  (by  volume)  sulphuric  acid. 

Close  the  cock  while  there  is  still  some  acid  in  the  funnel  (note  1). 

Discard  this  acid,  and  again  pass  through  the  tube  100  cc.  of  2  per  cent 


FIG.  18. 


ZINC  AS  A  REDUCING  AGENT  39 

acid,  stopping  as  before.  Test  this  acid  with  a  drop  of  permanganate  to 
see  if  it  contains  any  oxidizable  substance,  (Fe++  or  H202).  A  single 
drop  should  color  it  permanently;  if  it  does  not,  repeat  the  washing. 

Now  take  the  iron  solution  to  be  tested,  add  to  it  about  2  per  cent 
(by  volume)  of  sulphuric  acid,  heat  to  60°  C.,  and  pass  through  the 
reductor,  at  a  rate  not  to  exceed  50  cc.  per  minute.  Follow,  without 
allowing  air  to  enter,  with  150  cc.  of  2  per  cent  acid  (previously  pre- 
pared), and  then  with  75  cc.  of  water,  leaving  a  small  amount  of  the  latter 
in  the  funnel.  The  iron  is  now  ready  for  titration  and  the  reductor  is 
ready  for  the  next  determination  (provided  this  determination  is  carried 
out  the  same  day).  Titrate  the  iron  solution  immediately  in  the  suction 
flask. 

Procedure. — To  show  the  quantitative  action  of  the  reductor,  take 
50  cc.  of  the  ferrous  iron  solution  of  Exp.  18,  add  5  cc.  of  6  N  sulphuric 
acid,  titrate  carefully  with  the  permanganate,  recording  the  amount  of 
the  latter  used,  and  then  pass  the  solution  of  ferric  iron  through  the 
reductor,  as  explained  above,  and  titrate  again.  The  amount  of  per- 
manganate used  in  both  titrations  should  be  exactly  the  same  (note  2). 

NOTES. — (1)  If  air  is  allowed  to  enter  the  apparatus,  H2O2  will  be  formed,  and  this 
reduces  permanganate. 

(2)  If  the  ferrous  salt  contained  a  trace  of  ferric  the  amount  of  permanganate 
used  in  the  second  titration  will  slightly  exceed  that  used  in  the  first. 


CHAPTER  VIII 
SOLUBILITY  AND   SUPERSATURATION :     CONCENTRATION 

Exp.  20.  Supersaturated  Solutions  of  the  Hydrates  of  Sodium  Sulphate. 
Apparatus. — A  Biichner  funnel  (9  cm.)  set  up  according  to 
Fig.  19.  This  funnel  has  a  perforated  bottom  over  which  a  filter  paper 
is  placed.  It  is  always  used  with  suction,  like  a  Gooch  crucible,  and  is 
thus  a  great  time-saver  when  large  volumes  of  liquids  are  to  be  filtered. 
If  the  volume  of  liquid  is  small  it  is  well  to  receive 
it  in  a  large  test-tube  placed  inside  the  suction 
flask. 

Procedure. — Place  25  gm.  of  anhydrous  sodium 
sulphate  in  a  casserole,  add  40  cc.  of  water,  and 
heat  gently,  stirring  constantly  with  a  thermometer, 
until  the  temperature  reaches  about  40°  C.  Main- 
tain this  temperature  until  most  of  the  salt  dis- 
solves, and  any  possible  traces  of  the  hydrated 
phases  are  destroyed.  The  small  amount  of  pow- 
dered material  left  over  is  the  anhydrous  salt. 
(Would  this  dissolve  if  the  temperature  were  raised?) 
FIG  19.  Filter  immediately  on  a  Biichner,  making  sure 

beforehand  that  the  apparatus  is  perfectly  free  from 
any  traces  of  the  hydrated  salts,  or  any  salt,  in  fact.  When  this  is  done, 
return  the  solution  to  the  casserole  (carefully  rinsed)  and  again  heat  to 
40°  C.  (not  higher).  This  temperature  insures  absolute  freedom  from 
either  hydrate.  (Why?)  Now  transfer  the  solution  to  a  small  flask 
which  has  been  freshly  rinsed  with  distilled  water,  pouring  the  solution 
in  through  a  clean  funnel,  and  taking  great  care  not  to  get  a  trace  on  the 
neck  or  sides  of  the  flask.  Now  immediately  stopper  the  flask  with  a  wad 
of  clean  cotton,  and  then  set  it  in  an  ice-box,  or  any  place  where  the 
temperature  is  about  10°  C.  Be  careful  not  to  shake  the  flask  in 
moving  it. 

If  the  experiment  is  successful,  the  solution  will,  after  a  time,  deposit 
a  compact  mass  of  heptahydrate  crystals.  When  the  process  is  finished 
the  solution  is  in  equilibrium  with  the  heptahydrate,  but  is  strongly 
supersaturated  with  respect  to  the  decahydrate.  The  latter  fact  may 

40 


THE  TEST  FOR  POTASSIUM  41 

be  shown  by  dropping  in  a  minute  crystal  of  the  latter  salt.  When  this 
is  done  crystallization  will  begin  all  over  again.  Note,  however,  how 
different  is  the  shape  of  the  new  crystals  which  are  formed. 

Explain  each  part  of  the  procedure  in  this  experiment.  This  is  best 
done  by  drawing  the  solubility  curves,  and  making  them  the  basis  of  your 
reasoning. 

Exp.  21.  The  Test  for  Potassium. 

Apparatus. — A  test-tube  rack  and  10  test-tubes. 

Procedure. — Dissolve  10  gm.  of  tartaric  acid  and  10  gm.  of  potassium 
nitrate  each  in  50  cc.  of  water,  and  filter  the  two  solutions.  Place  a  row 
of  10  test-tubes  on  the  rack,  and  add  to  each  5  cc.  of  the  tartaric  acid 
solution.  Now,  to  tube  No.  1,  add  1  drop  of  the  potassium  nitrate 
solution,  to  No.  2  add  2  drops,  to  No.  3,  3  drops,  etc.  Gently  shake 
each  tube  to  mix  the  contents.  Is  a  precipitate  formed  in  any  case? 
With  a  glass  rod,  scratch  the  inside  of  each  tube,  beginning  with  No.  1, 
and  after  a  few  minutes,  notice  whether  anything  has  occurred.  If  you 
are  in  doubt  about  some  of  the  tubes  inoculate  the  contents  with  a 
minute  drop  from  No.  10.  In  which  tubes  were  the  solutions  super- 
saturated? 

Would  you  consider  this  a  fairly  delicate  test  for  potassium?  Un- 
der what  conditions  might  the  test  fail? 

Exp.  22.  Normal  Solutions  of  Hydrochloric  Nitric  and  Sulphuric  Acids. 

Apparatus. — An  accurate  hydrometer  having  a  range  of  1-1.2.  A 
tall  cylinder  to  use  with  the  hydrometer.  A  50-cc.  volumetric  flask, 
accurately  calibrated.  This  apparatus  is  calibrated  to  be  correct  at 
15°  C.  (60°  F.),  and  is  not  to  be  relied  upon  at  any  other  temperature. 

Procedure. — We  shall  first  determine  the  density  of  the  somewhat 
diluted  acids,  using  two  different  methods.  We  can  then  find  from  the 
accompanying  table,  the  actual  amount  of  acid  present  per  cubic  centi- 
meter, and  by  a  simple  calculation  can  determine  the  amount  needed 
for  500  cc.  or  1  liter  of  normal  concentration.  The  diluted  acids  are 
used  because  the  concentrated  acids  are  much  less  convenient  to  work 
with.  A  convenient  dilution  is  about  6  N.  Concentrated  hydro- 
chloric acid,  is  about  12  N,  concentrated  nitric  about  16  N,  and  con- 
centrated sulphuric  about  36  N.  To  prepare  the  6  N  acids  we  need, 
therefore,  to  dilute  6  cc.  of  hydrochloric  to  12  cc.,  6  cc.  of  nitric  to 
16  cc.,  and  6  cc.  of  sulphuric  to  36  cc.,  or  larger  amounts  in  like  propor- 
tions.* 

*  These  acids  may  be  made  up  in  large  amounts  for  the  class,  and  kept  where  the 
temperature  does  not  rise  above  15°  C. 


42        SOLUBILITY  AND  SUPERSATURATION:  CONCENTRATION 

It  is  understood,  of  course,  that  acids  made  up  by  specific  gravity, 
as  outlined  above,  will  be  accurate  only  within  about  1  per  cent;  but 
the  method  has  the  advantage  of  being  quick  and  easy,  and  for  many 
purposes  is  accurate  enough.  When  necessity  arises  we  shall  use  a 
more  accurate  method. 

Determine  the  density  by  both  of  the  following  methods,  two  or 
three  students  working  together : 

(a)  Take  enough  of  one  of  the  6  N  acids  to  fill  the  hydrometer 
cylinder  within  about  5  cm.  of  the  top,  adjust  the  temperature  to  15°  C. 
before  placing  in  the  cylinder,  and  then  measure  the  density  as  accurately 
as  possible  by  means  of  the  hydrometer.  The  readings  obtained  by  all 
the  groups  working  during  the  same  period  will  be  posted  on  the  black- 
board, and  the  average  of  all  these  readings  will  be  used  as  the  most 
probable  value. 

(6)  Clean  and  dry  the  volumetric  flask,  using  the  methods  of  Exp. 
10,  note  1.  Having  done  this,  weigh  the  flask  to  the  second  decimal 
place  only.  Now  again  adjust  the  temperature  of  the  acid  to  15°  C., 
and  then  fill  the  flask  with  it  at  this  temperature,  taking  care  that  the 
lower  edge  of  the  meniscus  comes  exactly  on  the  mark,  and  that  there 
are  no  drops  clinging  to  the  glass  above.  Weigh  the  flask  and  contents 
as  directed  in  the  case  of  the  empty  flask.  The  difference  between  these 
two  weights  constitutes  what  is  usually  termed  the  "  apparent  weight," 
or,  "  weight  in  air,"  of  the  acid.  The  true  weight,  or  weight  in  vacuum 
would  be  slightly  greater.  Thus,  on  an  object  having  a  density  of  1.1 
this  vacuum  correction  is  0.00095  gm.  per  gram  apparent  weight; 
where  the  density  is  1.2,  this  correction  is  0.00086  gm.  per  gram  apparent 
weight.*  Having  made  the  vacuum  correction  and  thus  obtained  the 
true  weight,  you  have  the  data  for  calculating  the  density,  which  is  the 
weight  of  1  cc.  The  readings  obtained  by  this  method  will  be  posted  as 
under  (a),  and  the  average  value  used  by  all  the  class. 

You  now  have  two  average  values  for  the  density  of  one  acid  obtained 
as  above,  and  may  take  for  your  working  value  the  mean  of  these  two. 
All  the  class  will  then  be  using  the  same  value,  and  must,  of  course, 
obtain  the  same  results. 

Now,  knowing  the  density  of  the  acid,  consult  the  following  tables 
and  determine  the  weight  of  actual  acid  per  cubic  centimeter  at  the 
determined  density.  Values  lying  between  those  given  must  be  found 
by  interpolation: 

*  For  a  wider  range  of  values,  see  Findlay,  Practical  Physical  Chemistry,  p.  29 
(1915). 


SOLUTIONS  OF  HYDROCHLORIC,  NITRIC  AND  SULPHURIC  ACIDS      43 


Hydrochloric  Acid  at  15°  C. 

Nitric  Acid  at  15°  C. 

Density. 

Grams  per  cc. 

Density. 

Grams  per  cc. 

1.070 

0.152 

1.155 

0.296 

1.075 

0.163 

1.160 

0.306 

1.080 

0.174 

1.165 

0.316 

1.085 

0.186 

1.170 

0.326 

1.090 

0.197 

1.175 

0.336 

1.095 

0.209 

1.180 

0.347 

1.100 

0.220 

1.185 

0.357 

1.105 

0.232 

1.190 

0.367 

1.110 

0.243 

1  .  195 

0.378 

1.115 

0.255 

1.200 

0.388 

Sulphuric  Acid  at  15°  C. 

Density. 

Grams  per  cc. 

1  .  155 

0.248 

1.160 

0.257 

1  .  165 

0.266 

1.170 

0.275 

1.175 

0.283 

1.180 

0.293 

1.185 

0.301 

1.190 

0.310 

1.195 

0.319 

1.200 

0.328 

Having  determined  the  weight  of  acid  per  cubic  centimeter,  calculate 
the  number  of  cubic  centimeters  required  for  500  cc.  of  normal  concen- 
tration, measure  out  the  amount  (still  at  15°  C.)  into  a  graduated  flask, 
using  a  burette  which  has  been  carefully  rinsed  with  the  acid,  and  make 
up  to  the  mark  with  distilled  water.  Mix  by  inverting  several  times — 
this  is  very  important. 

When  the  calculation  has  been  made  as  to  the  number  of  cubic  centi- 
meters of  6  N  acid  required,  it  is  a  wise  precaution  to  have  the  result 
checked  by  the  instructor  before  measuring  out  the  acid. 

After  one  acid  has  been  made  up,  determine  the  density  of  the  other 
two  by  the  same  method,  and  then  make  up  500  cc.  of  each  to  normal 
concentration.  Do  not  forget  the  precautions  about  temperature,  etc. 
In  determining  the  density  by  method  (6)  it  will  not  be  necessary  to  dry 


44        SOLUBILITY  AND  SUPERS ATURATION :  CONCENTRATION 

out  the  flask  and  weigh  again.  Simply  rinse  with  a  small  amount  of  the 
acid  you  intend  to  use,  and  then  fill  and  weigh. 

Any  unused  6  N  acid  may  be  returned  to  the  stock  bottles,  but 
care  should  be  taken  not  to  mix  the  different  acids. 

The  normal  acids  are  to  be  placed  in  the  600-cc.  bottles  of  the  outfit 
and  kept  for  future  use. 

Exp.  23.  Normal  Sodium  Hydroxide. 

Calculate  the  weight  of  actual  NaOH  required  for  600  cc.  normal 
alkali,  then  weigh  out  10  per  cent  more  than  this  to  allow  for  water  in 
the  sticks,  dissolve  in  water,  and  make  up  to  volume.  Mix  well,  as 
directed  under  22. 

Since  the  amount  of  water  in  the  sticks  is  not  accurately  known, 
you  must  determine  the  actual  concentration  of  this  solution  by  com- 
parison with  one  of  the  acids.  To  do  this  fill  one  of  the  burettes  (the 
rubber-tipped  one)  with  the  alkali,  and  the  other  with  the  normal  HC1, 
not  forgetting  the  proper  precautions  about  the  use  of  burettes.  Now 
run  off  about  25  cc.  of  the  acid  into  a  beaker;  add  2  drops  of  methyl 
orange  indicator,  and  then  titrate  with  the  alkali  to  the  first  appearance 
of  a  permanent  yellow  color.  Titrate  back  with  the  acid  to  the  appear- 
ance of  a  faint  pink.  The  true  end  point  is  a  combination  of  pink  and 
yellow — a  salmon  color — which  may  be  obtained  by  touching  off  from 
the  burette  tip  a  fraction  of  a  drop  of  either  solution,  as  required. 
When  the  titration  is  done,  carefully  take  the  burette  readings.  The 
amount  of  base  used  should  be  less  than  that  of  the  acid,  indicating  that 
it  is  more  concentrated.  The  readings  may  be  like  this:  NaOH  23.2  cc., 
HC1  25.3  cc.  This  indicates  that  the  alkali  is  more  concentrated  than 
the  acid  in  the  proportion  of  25.3  :  23.2.  To  be  reduced  to  the  same 
concentration  as  that  of  the  acid  it  must  be  diluted  in  just  the  proportion 
indicated  by  these  figures.  Calculate  from  your  figures  to  what  volume 
500  cc.  of  alkali  must  be  diluted,  and  then  measure  out  500  cc.  in  a 
graduated  flask,  transfer  to  the  600-cc.  bottle,  and  make  up  to  proper 
volume  by  adding  water  from  a  pipette  or  burette.  Mix  well  as  directed 
above,  and  then  check  by  retitrating  against  the  acid.  It  should  cor- 
respond exactly. 

When  the  sodium  hydroxide  is  made  to  correspond  with  one  of 
the  acids  as  above,  check  the  other  two  acids  by  titrating  against  it. 
They  may  differ  by  as  much  as  0.2  cc.  on  25  cc. 

The  solution  of  sodium  hydroxide  should  be  kept  in  a  bottle  with  a 
rubber  stopper;  a  glass  stopper  will  soon  become  hopelessly  set. 


CHAPTER   IX 


FREEZING-POINTS 


AND    BOILING-POINTS 
OSMOTIC  PRESSURE 


OF   SOLUTIONS: 


J 


Exp.  24.  Molecular  Lowering  of  the  Freezing-point  of  Water. 

Apparatus. — A  thermometer  graduated  in  0.1°  and  having  the 
range  —10  to  +50°;  a  strong  test-tube,  25X180  mm.;  a  larger  tube 
to  serve  as  a  jacket;  an  aluminum  stirrer;  a  small  battery  jar  (4  X4  in.) 
to  serve  as  a  cooling  bath.  The  apparatus  is  assembled  as  seen  in 
Fig.  20,  the  jacket  tube  being  supported  by 
a  ring  stand,  which  also  holds  the  battery 
jar: 

Procedure. — Make  up  a  solution  of  ethyl 
alcohol  containing  1  mole  of  actual  C2H50H 
in  1000  gm.  of  water.  Use  the  so-called 
"  absolute  alcohol  "  of  99.9  per  cent  purity, 
and  make  up  not  to  exceed  100  cc.  of  the 
solution.  Since  alcohol  is  a  volatile  liquid 
and  also  likely  to  absorb  moisture  from  the 
air,  the  best  procedure  for  weighing  will 
probably  be  first  to  weigh  accurately  (to  the 
second  place)  a  small  stoppered  flask  which 
has  previously  been  thoroughly  dried,  then  . 
measure  into  it  by  means  of  a  dry  pipette 
5  cc.  of  the  alcohol,  and  weigh  again.  Knowing 

the  weight  of  the  alcohol,  it  is  an  easy  matter  to  calculate  what  weight 
of  water  must  be  added  to  give  the  proportions  above  mentioned. 
Having  made  this  calculation,  add  to  the  alcohol  the  proper  amount  of 
water,  and  mix  thoroughly. 

Prepare  a  freezing  bath  by  mixing  finely  crushed  ice  and  water,  and 
then  adding  enough  common  salt  to  lower  the  temperature  to  —6°  C. 
Place  in  the  battery  jar  such  a  quantity  of  the  mixture  that,  when  the 
jacket  is  immersed  about  3  in.  deep,  the  jar  is  nearly  full. 

First  determine  the  freezing-point  of  water  alone  (note  1),  proceeding 
as  follows:  Place  about  20  cc.  of  water  in  the  inner  tube,  insert  the 
thermometer  and  stirrer  as  shown  in  the  sketch,  and  then  cool  by  insert- 

45 


FIG.  20. 


46          FREEZING-POINTS  AND  BOILING-POINTS  OF  SOLUTIONS 

ing  directly  in  the  bath,  stirring  constantly.  When  the  temperature  has 
fallen  to  about  —1°  or  —  2°,  stir  vigorously  to  induce  freezing  (note  2). 
When  ice  begins  to  separate,  place  the  tube  in  the  jacket  and  stir  gently 
for  a  moment  until  the  temperature  becomes  constant;  then  read  the 
thermometer  with  a  lens,  tapping  gently,  and  estimating  to  0.01°.  Be 
sure  that  numerous  ice  crystals  are  floating  about  in  the  liquid  when  the 
reading  is  taken,  or  the  temperature  may  not  be  the  freezing-point 
at  all  (note  3).  Now  remove  the  inner  tube,  thaw  the  ice  by  holding 
the  tube  in  the  hand  for  a  moment,  and  then  repeat  the  determination. 
At  least  five  readings  should  be  taken  in  this  way.  If  these  do  not  differ 
by  more  than  0.03°,  the  average  may  be  taken  as  the  freezing-point  of 
the  water. 

Determine  the  freezing-point  of  the  alcohol  solution  as  follows: 
Remove  the  water  from  the  inner  tube,  rinse  the  latter  with  the  alcohol 
solution,  and  then  add  about  20  cc.  of  the  same.  Cool  in  the  bath  as 
above,  until  a  temperature  of  —4°  or  —5°  is  reached,  and  then  to  induce 
freezing  inoculate  with  a  fragment  of  clean  ice  the  size  of  a  wheat  grain. 
Remove  the  tube  from  the  bath,  and  stir  until  the  amount  of  ice  has 
diminished  somewhat  (note  4),  then  place  in  the  jacket,  and  after  stirring 
a  moment,  read  the  temperature.  Do  not  forget  that  you  are  reading 
downward  from  the  zero,  and  that,  therefore,  the  hundredths  are  to  be 
estimated  downwards  from  the  nearest  division  above  the  end  of  the 
mercury  thread.  Thaw  the  ice  and  repeat  as  above,  obtaining  five 
concordant  readings.  Take  the  average  jf  these  as  the  freezing-point  cf 
the  solution. 

The  molecular  lowering  for  water  is  the  difference  between  the  freez- 
ing-point of  the  pure  water  and  that  of  the  solution.  Calculate  this. 

NCTES. — (1)  Water  freezes  at  zero,  of  course;  but  the  thermometer  may  not  be 
correct. 

(2)  Pure  water  tends  to  become  supercooled  without  freezing  just  as  a  solution 
becomes  supersaturated  without  depositing  crystals,  and  the  same  means  have  to  be 
taken  to  prevent  this  condition. 

(3)  If  no  ice  is  present  the  temperature  may  be  either  above  or  below  the  freezing- 
point.     If  numerous  ice  crystals  are  present  and  the  water  is  kept  stirred  so  as  to 
bring  these  rapidly  into  contact  with  all  parts  of  the  liquid,  the  temperature  will 
remain  at  the  freezing-point  until  all  the  water  has  solidified.     Abstraction  of  heat 
does  not  change  the  temperature  but  simply  hastens  the  change  of  state.     The  same 
sort  of  thing  is  true  when  the  ice  is  melting;  the  change  of  state  here  absorbs  heat,  so  if 
the  amount  of  ice  is  fairly  large  and  good  contact  between  ice  and  water  is  kept  up  by 
stirring,  the  temperature  remains  at  0°.     Too  rapid  heating,  with  too  small  an  amount 
of  ice  present  may  result  in  raising  the  temperature  above  0°.     It  is  merely  a  case  of 
equilibrium  between  the  rates  at  which  heat  can  be  delivered  and  absorbed. 

(4)  When  a  non-saturated  solution  freezes,  ice  alone  separates.     This  makes  the 
solution  more  concentrated  and  thus  makes  the  freezing-point  lowering  greater  than 


MOLECULAR  WEIGHT  OF  PROPYL  ALCOHOL 


47 


is  expected.  We  are,  therefore,  careful  not  to  have  too  large  an  amount  of  ice 
separated  at  the  time  we  take  the  reading.  However,  the  amount  must  not  be  re- 
duced to  a  few  minute  crystals,  or  the  temperature  may  rise  above  the  freezing-point. 

Exp.  25.  Molecular  Weight  of  Propyl  Alcohol. 

Apparatus. — Same  as  that  of  Exp.  24. 

Procedure. — Make  up  a  solution  containing  propyl  alcohol  in  some 
known  amount,  say  100  gm.,  per  1000  gm.  of  water.  Ten  gm.  dissolved 
in  100  gm.  of  water  will  be  enough.  Determine  the  freezing-point  of  the 
solution  exactly  as  directed  in  Exp.  24.  If  Exp.  24  has  not  been  per- 
formed it  will  be  necessary  to  carry  out  the  first  part  here  also,  namely, 
the  determination  of  the  freezing-point  of  the  pure  water  on  the  thermo- 
meter you  are  using.  The  molecular  lowering  for  water,  1.86°,  may 
be  used  in  the  calculation  below  without  determination. 

When  the  freezing-point  of  water  and  that  of  the  solution  have  been 
determined,  calculate  from  your  data  what  weight  of  propjd  alcohol 
would  have  been  necessary  to  give  the  molecular  lowering;  this  is  its 
molecular  weight.  The  formula  for  propyl  alcohol  is  CsHjOH.  How 
nearly  correct  is  your  value? 

Exp.  26.  Qualitative  Experiment  on  Osmotic  Pressure. 

Apparatus. — A  diffusion  thimble  of  parchment  paper  A  Fig.  21;  a 
glass  adapter  B  of  the  proper  size  to  fit  the  thimble 
after  the  latter  has  been  thoroughly  soaked  in  boil- 
ing water;  a  capillary  tube  3  or  4  feet  long,  1-mrn. 
bore  C.  Prepare  the  apparatus  as  follows:  Fill  the 
thimble  with  water,  immerse  in  a  beaker  of  water  and 
boil  until  expanded  and  free  from  air;  then  draw  over 
the  end  of  the  adapter  and  tie  securely  in  place  with 
waxed  thread.  When  not  in  use  the  thimble  and 
adapter  should  be  kept  in  water  to  prevent  drying  out. 

Procedure. — Prepare  a  concentrated  solution  of 
cane  sugar.  Fill  the  thimble  by  pouring  the  solution 
through  a  slender  funnel.  The  rubber  connector 
should  be  in  place,  and  this  should  be  filled  also. 
Finally,  holding  the  apparatus  by  the  adapter,  push 
into  connection  with  the  capillary  tube.  Some  of  the 
solution  will,  of  course,  be  forced  a  short  distance  up 

the    tube,  but    this    does    no    harm.      Suspend    the       

thimble  now  in  a  cylinder  of  water  as  shown  in  the          pIG  21 
sketch.     If  the  apparatus  is  tight   and   properly  set 
up,  the  solution  should  ascend  the  tube  at  the  rate  of  about  half  a  centi- 
meter per  minute. 


CHAPTER  X 
THE  THEORY  OF  IONIZATION 

Exp.  27.  Salt  Effect. 

Procedure  A.  Salt  Effect  with  Acetic  Acid.— To  300  cc.  of  distilled 
water  add  6  drops  of  methyl  orange  indicator,  stir  thoroughly,  and  then 
remove  100  cc.  to  another  beaker.  To  the  remaining  200  cc.  now  add 
sufficient  N  acetic  acid  to  give  the  characteristic  salmon  color  of  the 
end  point  (about  3  drops).  After  stirring  thoroughly,  divide  into  two 
equal  parts.  To  one  of  these  portions  containing  the  acetic  acid  add 
about  10  gm.  of  pure,  neutral  sodium  chloride.  Stir  until  solution  is 
complete,  and  then  place  alongside  the  second  acidulated  portion  for 
comparison.  Is  there  any  evidence  of  greater  ionization,  resulting  in 
stronger  acidity? 

To  prove  that  the  change  is  not  due  to  any  action  of  the  salt  on  the 
indicator  itself,  add  a  similar  amount  of  the  sodium  chloride  to  the  por- 
tion containing  the  indicator  alone.  If  the  salt  is  really  neutral  there 
should  be  no  change  in  color. 

Show  why  water  causes  acetic  acid  to  ionize  and  why  sodium  chloride 
enhances  the  effect. 

Procedure  B.  Salt  Effect  with  Ammonium  Hydroxide. — To  50  cc. 
of  a  M/5  solution  of  magnesium  sulphate  add  10  cc.  of  N  ammonium 
hydroxide,  stir,  and  let  stand  for  five  minutes.  Some  of  the  magnesium 
will  be  precipitated  as  the  hydroxide,  Mg(OH)2,  by  the  hydroxyl  ion 
from  the  base;  but,  due  to  slight  ionization  of  the  latter  (note  1)  the 
precipitation  is  not  complete.  Anything  that  will  increase  the  ionization 
of  the  ammonium  hydroxide  will  cause  further  precipitation  of  the  mag- 
nesium. To  show  that  sodium  chloride  does  this,  filter  the  solution  on  a 
Biichner  funnel,  and  then,  to  the  clear  filtrate,  add  about  10  gm.  of  pure, 
neutral  salt.  Explain  the  result. 

NOTE. — (1)  The  ammonium  salt  formed  in  the  reaction  also  suppresses  the 
ionization  of  the  ammonium  hydroxide,  but  this  in  no  Way  interferes  with  the  accu- 
racy of  the  experiment. 

Exp.  28.  Degree  of  Ionization  from  Abnormal  Freezing-point  Lowering. 
Apparatus. — Same  as  that  used  in  Exp.  24. 
Procedure. — Make  up  about  100  cc.  of  a  solution  of  sodium  chloride 

48 


IONIZATION  AND  CHEMICAL  TESTS  49 

containing  the  salt  at  the  rate  of  1  mole  in  1000  gm.  of  water.  Deter- 
mine the  freezing-point  lowering  as  in  the  case  of  propyl  alcohol,  obtain- 
ing five  values  as  there  directed. 

When  the  data  have  been  obtained,  the  degree  of  ionization  may  be 
calculated  as  follows:  Let  a  equal  the  fraction  ionized  (the  degree  of 
ionization),  and  let  n  equal  the  number  of  ions  resulting  from  the  disso- 
ciation of  1  molecule.  Also  let  d  equal  the  observed  molecular  lowering. 
Now  if  the  breaking  up  of  1  molecule  produces  n  ions,  we  obtain  in  the 
case  of  this  1  molecule  n—l  particles  more  than  we  had  to  begin  with. 
If  a  is  the  fraction  of  molecules  ionized,  the  total  number  of  particles 
added  by  this  process  will  be  a  (n—l);  and  if  1  equals  the  number  of 
molecules  before  any  are  broken  up,  then  the  total  number  of  particles 
now  present  will  be  l+a  (n—l).  Finally,  since  the  freezing-point 
lowering  is  proportional  to  the  number  of  particles  present,  no  matter 
what  their  size,  the  observed  lowering  will  be  related  to  the  true  molecular 
lowering  (1.86°)  just  as  the  total  number  of  particles  is  related  to  the 
original  number.  This  may  be  expressed  thus: 

d  :  1.86::l+a(n-n  :  1 
From  this  we  obtain 

d-1.86 
1.86  (n-1) 

Substitute  your  data  in  this  formula,  and  calculate  the  degree  of 
ionization  of  the  salt  you  used  (note  1). 

NOTE. — (1)  The  value  will  be  somewhat  too  large  on  account  of  the  hydration  of 
the  ions. 

Exp.  29.  Ionization  and  Chemical  Tests. 

Procedure  A.  Ionization  and  Speed  of  Displacement  of  Hydrogen. — 

Take  three  test-tubes:  fill  No.  1  two-thirds  full  of  N  hydrochloric  acid, 
No.  2  with  N  phosphoric  acid  (note  1),  and  No.  3  with  N  acetic  (note  2). 
Take  three  pieces  of  pure  sheet  zinc  (1  cm.  X2  cm.)  and  wrap  each  neatly 
in  a  small  piece  of  brass  gauze  (note  3).  Having  done  this,  drop  them 
into  the  test-tubes,  and  after  waiting  for  the  reaction  to  get  well  under 
way,  note  the  relative  speeds.  The  reaction,  as  you  know,  consists  in 
the  displacement  of  hydrogen  by  zinc.  Consult  the  table  of  ionization, 
and  note  whether  the  speeds  are  proportional  to  the  degrees  of  ionization 
of  the  individual  acids. 

Procedure  B.  Ionization  of  Iron  Salts  and  Iron  Complexes. — Treat  a 
solution  of  ferric  chloride  with  a  few  drops  of  a  solution  of  ammonium  or 
potassium  thiocyanate.  Repeat  the  same  test,  using  potassium  ferri- 


50  THE  THEORY  OF  IONIZATION 

cyanide  as  the  carrier  of  ferric  iron.  Explain  the  difference  between  the 
two  tests. 

To  one  5-cc.  portion  of  potassium  ferric-oxalate  solution  add  ammo- 
nium thiocyanate,  and  to  another  add  sodium  hydroxide.  Why  do  you 
obtain  a  test  for  ferric  ion  in  one  case  and  not  in  the  other? 

Procedure  C.  Chloride  and  Chlorate  Ion. — To  5  cc.  each  of  solutions 
of  sodium  chloride  and  potassium  chlorate  add  a  few  drops  of  silver 
nitrate  solution.  Explain  the  results. 

Procedure  D.  lonization  of  Boric  Acid. — Test  a  solution  of  boric  acid 
with  a  drop  of  methyl  orange.  Is  enough  H+  ion  present  to  affect  the 
indicator?  Why? 

Procedure  E.  lonization  of  Mercury  Salts. — To  separate  5-cc.  por- 
tions of  mercuric  chloride  and  mercuric  nitrate  solutions  add  a  solution 
of  potassium  dichromate.  Look  up  the  degrees  of  ionization  of  these 
two  salts  of  mercury,  and  then  decide  whether  the  tests  obtained  are  in 
accord  with  the  ionization  data. 

NOTES. — (1)  Phosphoric  acid  acts  practically  like  a  monobasic  acid,  the  second- 
ary and  tertiary  ionization  being  so  slight.  "  Normal  "  may,  therefore,  be  taken  to 
mean  "  molar,"  if  methyl  orange  indicator  is  used  in  making  it  up.  This  acid  is  best 
made  up  by  the  specific  gravity  method  of  Exp.  22.  The  concentrated  acid  has  a 
density  of  1.7,  and  contains  85  per  cent  actual  H3PO4. 

(2)  Glacial  acetic  acid  is  approximately  100  per  cent  pure,  and  the  density  is 
1.055.     The  normal  acid  is  best  made  up  in  this  case,  also,  by  the  method  of  Exp.  22. 

(3)  Pure  zinc  reacts  very  slowly  with  an  acid.     With  the  gauze  the  zinc  forms  an 
electric  couple,  causing  the  reaction  to  proceed  smoothly,  but  still  in  proportion  to 
the  ionization  of  the  several  acids. 

Exp.  30.  Ionization  and  Catalysis. 

Procedure  A.  Acid  Catalysis. — Take  four  100-cc.  flasks  and  place  in 
each  40  cc.  of  water.  To  the  four  in  order  then  add  an  accurate  10-cc. 
portion  of  one  of  the  following  acids,  viz.,  N  HC1,  N  HNOs,  N  H^SCU, 
and  N  H3PO4  (note  1).  That  is,  add  HC1  to  No.  1,  HN03  to  No.  2,  etc. 
Now  to  each  flask  add  exactly  2  cc.  of  methyl  acetate,  and  mix  the  con- 
tents thoroughly  by  closing  with  a  cork  and  shaking.  Let  the  reaction 
proceed  for  fifteen  hours  (not  longer) . 

In  the  meantime,  measure  out  with  the  same  pipette  10-cc.  portions 
of  each  of  the  above  acids,  and  determine  by  titration  the  exact  number 
of  cubic  centimeters  of  N  NaOH  required  to  neutralize  them,  using 
phenolphthalein  as  indicator  (note  2). 

At  the  end  of  the  fifteen  hours,  add  to  each  flask  from  a  burette  the 
proper  amount  of  N  NaOH  to  neutralize  the  acid  originally  added; 
and  then,  using  phenolphthalein  indicator,  titrate  the  acetic  acid  pro- 
duced by  the  hydrolysis  in  each  case.  On  account  of  the  excess  of 
methyl  acetate  always  present  here,  the  end-point  color  of  the  indicator 


IONIZATION  AND  CATALYSIS 


51 


will  not  remain  long  after  the  titration  is  finished.  (Explain  in  light  of 
Procedure  B.)  Take  as  the  end  of  the  titration  the  point  where  a  faint 
pink  color  remains  for  ten  seconds  after  stirring. 

Remember  that  the  number  of  cubic  centimeters  of  alkali  used  repre- 
sent quantities  of  acetic  acid,  and  that  quantities  of  acetic  acid  represent 
the  speeds  of  the  hydrolytic  reactions.  With  this  in  mind,  arrange  the 
several  numbers  of  cubic  centimeters  in  descending  order,  placing  them  in 
column  2  of  the  table  following.  In  column  1  write  the  names  of  the 
corresponding  catalyzing  acids,  and  in  column  3,  the  degree  of  ionization 
of  these  acids  as  obtained  from  the  table  in  the  text. 


Catalyzing  Acids. 

Cc.  of  N  H  LIC 
Produced. 

Ionization  of  the 
Catalyzing  Acids. 

Do  you  note  a  direct  proportionality  between  ionization  and  catal- 
ysis? (note  3). 

Procedure  B.  Basic  Catalysis. — Into  each  of  two  beakers  put  50  cc. 
of  water  and  1  drop  of  phenolphthalein.  Then  add  to  one  beaker  1  cc. 
of  N  NaOH  and  to  the  other  1  cc.  of  N  NH4OH.  Finally  add  to  each 
(to  the  ammonia  first),  2  cc.  of  methyl  acetate,  quickly  mix  by  stirring, 
and  note  the  time.  The  acetic  acid  generated  by  the  reaction  neutralizes 
the  alkali,  and  thus  removes  the  color  of  the  indicator.  The  time 
required  is  inversely  proportional  to  the  ionization  of  the  alkali  in  the 
two  cases. 

NOTES. — (1)  This  is  really  molar  phosphoric  acid;  that  is,  normal  to  methyl 
orange. 

(2)  Phenolphthalein  is  used  here  because  it  must  be  used  later  in  titrating  the 
acetic  acid  set  free  in  the  reaction.     With  this  indicator  phosphoric  acid  will  act  as  a 
dibasic  acid,  and  so  will  appear  to  be  of  2  N  concentration. 

(3)  The  hydrogen  ion  is  not  used  up  in  the  process  of  catalysis,  but  the  methyl 
acetate  is.     For  this  reason  the  reaction  tends  to  slow  down  as  it  proceeds,  and  finally 
come  to  a  standstill.     This  effect  is  less  noticeable  in  the  case  of  a  weak  acid,  where 
the  reaction  is  slow,  because  the  amount  of  acetate  remains  nearly  constant.     Hence 
the  final  result  will  probably  be  somewhat  to  the  advantage  of  the  weaker  acids. 
In  other  words,  it  will  make  the  weaker  acids  appear  a  little  stronger  than  they  are. 


52  THE  THEORY  OF  IONIZATION 

Exp.  31.  Heat  of  Neutralization. 

Apparatus. — The  calorimeter  apparatus  used  in  the  determination  of 
specific  heat.  The  beaker  should  hold  not  more  than  300  cc.,  allowing 
complete  submersion  of  the  thermometer  bulb  when  the  beaker  contains 
200  cc.  of  liquid.  The  thermometer  should  be  graduated  to  0.1°  and 
should  have  a  range  of  0°-50°. 

The  water  equivalent  of  the  calorimeter  may  be  calculated  from  its 
weight;  but  since  only  part  of  the  beaker  will  be  heated,  and  since  the 
bulb  of  the  thermometer  is  not  thus  included,  a  better  method  is  the 
following : 

Place  100  cc.  of  water  in  the  calorimeter  and  take  its  temperature 
accurately.  Take  another  100  cc.  of  water  and  warm  it  to  a  tem- 
perature about  8°  higher,  and  take  its  temperature  accurately  with 
the  same  thermometer.  Next  pour  this  second  portion  of  water  into  the 
calorimeter.  Proceed  then  to  measure  the  temperature  of  the  mixture, 
but  first  cool  the  thermometer  bulb  back  to  its  original  temperature 
by  immersing  in  another  beaker  of  water,  such  as  was  first  put  in  the 
calorimeter.  The  mixture  should  be  thoroughly  stirred  before  the  tem- 
perature is  read.  The  rise  in  the  temperature  of  the  cold  water,  mul- 
tiplied by  100,  will  be  the  heat  received  by  it.  The  fall  in  temperature 
of  the  warm  water,  multiplied  by  100  will  be  the  heat  lost  by  it.  The 
difference  between  these  products  will  be  the  heat  received  by  the  cal- 
orimeter and  thermometer.  This  amount,  divided  by  the  change  in 
temperature  of  the  apparatus  will  be  the  water  equivalent  of  the  latter 
in  grams. 

Procedure. — First  prepare  N/2  solutions  of  NaOH,  HC1  and  NH03 
by  diluting  the  normal  solutions  already  on  hand.  Proceed  then  to 
determine  the  heat  of  neutralization  of  the  two  acids  as  follows :  Measure 
out  exactly  100  cc.  of  the  NaOH  and  of  one  of  the  acids  and  place  in  150 
cc.  beakers  of  tall  form.  Set  these  on  a  clean  porcelain  tile,  and  suspend 
above  them  the  thermometer  in  such  a  position  that  it  reaches  almost 
to  the  bottom  of  either  beaker.  Measure  the  temperature  of  the  two 
solutions  very  accurately,  tapping  the  thermometer  and  waiting  until 
the  reading  is  constant.  Use  a  lens.  In  transferring  the  thermometer 
from  one  solution  to  the  other,  remove  the  adhering  drop  by  touch- 
ing the  bulb  to  the  side  of  the  beaker.  Record  the  average  of  the  two 
temperatures.  Now  set  the  tile,  carrying  the  two  beakers,  to  one  side, 
and  put  in  its  place  the  calorimeter,  adjusting  the  thermometer  so  that 
it  almost  touches  the  bottom  of  the  beaker.  Finally,  pour  the  two 
solutions  together  into  the  calorimeter  and  stir  with  the  thermometer. 
For  reading  the  temperature  make  use  of  the  following  scheme: 

Note  the  exact  time  when  the  two  solutions  are  mixed,  and  at  the 


HEAT  OF  NEUTRALIZATION 


53 


28°- 

o 

27-- 

26C 

25- 

24 
23.502, 


23 


end  of  exactly  one  minute  begin  a  series  of  readings  taken  one  minute 
apart  and  lasting  until  the  rate  of  fall  becomes  constant.  Plot  the  values 
on  coordinate  paper,  and  obtain  the  true  maximum  temperature  by 
extrapolation.  An  example  will  make  this  clear:  The  temperature 
before  mixing  was,  in  a  certain  case,  23.45°,  Beginning  one  minute 
after  mixing,  the  read- 
ings were  26.48,  26.53, 
26.48,  26.44,  26.40, 
and  26.36.  These 
values  are  seen  plotted 
in  Fig.  22. 

From  A  to  B  the 
slope  of  the  curve  is 
constant;   that  is,  the 
fall  in  temperature  is 
the  same  per  minute. 
If   this    curve    is    ex- 
tended   back     (extra-    — L 
polated)  to  the  Y-axis, 
it  marks  the  tempera- 
ture which  really  ex- 
isted at  the  moment  of  mixing,  and  which  would  have  been  indicated 
later  by  the  thermometer  if  no  heat  had  been  lost  by  radiation.     This 
temperature  is  26.61°.     The  real  rise  then,  due  to  the  neutralization, 
was  26.61-23.45°,  or  3.16°. 

When  the  final  temperature  has  been  read  and  the  rise  calculated  by 
use  of  the  above  scheme,  the  total  heat  evolved  is  calculated  in  the  usual 
way.  The  total  volume  of  the  solution  is  200  cc.  and  since  it  is  so  dilute 
(N/4),  it  may  be  reckoned  as  200  cc.  of  water.  To  this  must  be  added 
the  water  equivalent  of  the  calorimeter  and  thermometer  bulb  deter- 
mined experimentally  as  directed  above. 

.  When  the  total  amount  of  heat  produced  by  the  neutralization 
has  been  calculated,  it  must  be  remembered  that  only  a  fraction  of 
an  equivalent  was  taken.  It  then  remains  to  calculate  by  proportion 
what  amount  of  heat  would  have  been  produced  if  a  whole  equivalent 
weight  of  acid  had  been  neutralized.  This  will  be  the  heat  of  neutral- 
ization. It  should  not  differ  from  13,700  cal.  by  more  than  200  cal.  in 
the  case  of  either  acid. 

Explain  fully  the  significance  of  this  experiment. 


CHAPTER  XI 
INDICATORS 

Exp.  32.  Sensitiveness  of  Methyl  Orange  Indicator  and  its  End-point 

Correction. 

Apparatus. — Ten  Nessler's  tubes,  100  cc.;  a  color  comparator  as 
seen  in  Fig.  23.  The  lower  bar  A  of  the  comparator  is  double,  and 
between  the  two  parts  is  a  diaphragm  of  colorless  celluloid  upon  which 
the  Nessler's  tubes  rest.  The  light  is  reflected  up  through  the  tubes  from 

the  white  enameled  surface  of  the 
slanting  board  B.  When  in  use,  the 
comparator  is  placed  so  as  to  face  a 
large  window.  Artificial  light  can 
scarcely  be  used  for  color  work,  except- 
ing, possibly,  the  so-called  "  daylight 
lamps." 

Be  sure  that  the  Nessler's  tubes 
are  clean  and  free  from  the  least  trace 
of  acid  or  alkali.  The  Nessler's  tubes 
must  not  be  set  up  on  an  open 
desk;  they  should  always  be  placed 
in  the  comparator  to  avoid  breakage. 

Procedure. — Starting  with  your  normal  hydrochloric  acid,  prepare  1 
liter  of  N/1000  acid.  Place  seven  of  the  tubes  in  the  comparator. 
Beginning  then  at  one  end  of  the  row,  add  in  succession  to  the  tubes  the 
following  amounts  of  the  N/1000  acid:  70  cc.,  60  cc.,  50  cc.,  40  cc.,  30  cc., 
20  cc.,  10  cc.  Having  done  this,  fill  the  tubes  with  distilled  water  to  the 
100-cc.  mark.  Before  going  further  with  the  experiment  calculate  the 
concentration  of  the  acid  in  each  tube  and  state  it  both  as  a  decimal 
and  in  terms  of  the  minus  powers  of  10.  Also  submit  your  values  to  the 
instructor  for  verification. 

To  each  tube  now  add  two  drops  of  methyl  orange  indicator  *  and 
then  stir  the  contents  thoroughly  by  closing  with  the  hand  and  inverting. 


FIG.  23. 


*  0.1  gm.  of  the  solid  in  100  cc.  of  water  is  the  usual  concentration. 

54 


SENSITIVENESS  OF  PHENOLPHTHALEIN  INDICATOR         55 

Notice  which  concentration  gives  the  salmon  color  commonly  recognized 
as  the  end  point  for  this  indicator.  Notice  also  that  in  the  more  dilute 
solutions  the  concentration  of  the  H+  ion  is  too  small  to  affect  the 
indicator,  making  it  appear  as  though  the  solution  were  neutral,  which 
it  is  not. 

At  the  dilutions  here  used,  we  are  safe  in  saying  that  the  acid  is  all 
ionized  in  each  case.  If  this  is  so,  then  the  total  concentrations,  as  you 
worked  them  out  above,  represent  also  the  concentrations  of  the  hydro- 
gen ion.  Place  these  concentrations  in  a  horizontal  row,  labeling  them 
(H+),  meaning  "  hydrogen  ion  concentration."  Work  out  the  corre- 
sponding hydroxyl  ion  concentration  in  each  case  and  place  these 
below  the  others.  They  should  be  labeled  (OH~).  In  testing  the 
accuracy  of  these  latter  values  you  have  only  to  notice  whether  the 
product  (H+)  X  (OH~)  in  any  case  is  10~14. 

What  concentration  of  H+  and  OH~  are  present  in  a  solution  which 
gives  the  end-point  color  with  methyl  orange?  Check  this  point  under 
the  proper  concentrations  by  means  of  an  arrow,  j ,  and  label  it  "  methyl 
orange  end  point." 

End-point  Correction. — Suppose  you  were  titrating  a  solution  of 
hydrochloric  acid  in  a  volume  of  200  cc. ;  when  the  end  point  is  reached, 
what  weight  of  HC1  remains  unneutralized?  What  part  of  1  cc.  of 
N  HC1  does  this  amount  represent? 

If  you  placed  50  cc.  of  N  HC1  in  a  beaker,  diluted  it  to  400  cc.  and 
then  titrated  to  the  methyl  orange  end  point  with  NaOH,  what  per  cent 
of  the  acid  would  remain  unneutralized?  To  state  it  otherwise,  what 
would  be  the  percentage  error  in  the  titration? 

What  fraction  of  1  cc.  of  N/10  HC1  will  be  required  to  give  the 
methyl  orange  end  point  in  a  volume  of  500  cc.?  If,  then,  in  a  certain 
titration  we  used  49.72  cc.  of  N/10  HC1  and  the  total  volume  of  the 
solution  is  500  cc.,  what  volume  of  the  acid  was  actually  used  in  the 
neutralization? 

Exp.  33.  Sensitiveness  of  Phenolphthalein  Indicator. 

Apparatus. — Same  as  in  Exp.  32. 

Procedure. — Dilute  1  cc.  of  N  sodium  hydroxide  to  10  cc.  and  then 
dilute  1  cc.  of  this  solution  to  1000  cc.,  preparing  thus  a  solution  of  alkali 
having  a  concentration  of  0.0001.  Place  seven  Nessler's  tubes  on  the 
comparator,  and  add  to  them  in  succession  the  following  amounts  of  the 
above  solution:  100  cc.,  90  cc.,  80  cc.,  70  cc.,  60  cc.,  50  cc.,  and  40  cc. 
Having  done  this,  fill  the  tubes  to  the  100-cc.  mark  with  distilled 
water. 

Assuming  now  that  all  the  NaOH  is  ionized,  work  out  for  each  tube 


56  INDICATORS 

the  concentration  of  OH~  and  H+,  and  record,  as  directed  in  Exp.  32, 
in  terms  of  the  minus  powers  of  10. 

To  determine  which  of  these  concentrations  of  H+  and  OH~  pro- 
duces the  first  suggestion  of  a  pink  color  (the  "  end  point ")  with 
phenolphthalein,  add  to  each  tube  two  drops  of  the  indicator  solution,* 
and  mix  by  closing  with  the  hand  and  inverting  (note  1).  Check  this 
point  under  the  proper  concentrations  as  directed  in  32,  and  label  it 
"  Phenolphthalein  End  Point  "  (note  2). 

NOTES. — (1)  Be  very  sure  that  there  is  no  trace  of  acid  or  alkali  on  the  hand  here. 
Such  a  trace  might  amount  to  more  than  the  alkalinity  of  the  extremely  dilute  sol- 
tions  with  which  you  are  working. 

(2)  The  value  obtained  by  this  simple  dilution  method  will  not  be  so  nearly  cor- 
rect as  in  the  case  of  methyl  orange.  This  is  due  to  the  fact  that  a  considerable  pro- 
portion of  the  alkali  in  such  an  extremely  dilute  solution  is  neutralized  by  the  carbonic 
acid  (from  the  air)  always  present  in  distilled  water.  This  makes  the  color  change  of 
the  indicator  appear  to  come  where  the  OH~  concentration  is  greater  than  it  really  is. 
The  true  value  is  (OH~)  10  ~6  and  (H+)  10  ~8. 

Exp.  34.  Choice  of  an  Indicator.! 

Procedure  A.  Titration  of  a  Weak  Acid. — Measure  out  with  a  pipette 
10  cc.  of  N  acetic  acid,  and  titrate  with  N  sodium  hydroxide,  using 
methyl  orange  as  indicator.  Does  the  end  point  appear  when  the 
amounts  of  acid  and  base  are  equal?  Repeat  the  titration  with  another 
portion  of  acid,  using  phenolphthalein  as  indicator.  Explain  the 
results. 

Procedure  B.  Titration  of  a  Weak  Base. — Measure  out  10-cc. 
portions  of  N  ammonium  hydroxide,  dilute  to  about  50  cc.  to  reduce  the 
volatility,  and  then  titrate  with  N  hydrochloric  acid,  using  in  one  case 
methyl  orange  and  in  the  other  case  phenolphthalein,  as  indicator. 
Explain  the  results. 

Note  that  one  of  the  indicators  used  above  was  especially  sensitive 
to  OH~  ion,  the  other  to  H+  ion.  Can  you  make  a  general  statement 
as  to  the  kind  of  indicator  necessary,  (a)  with  weak  acids?  (6)  with 
weak  bases? 

Consult  the  indicator  table  below  and  select  two  indicators  which 
you  think  would  take  the  place  of  the  two  used  above  for  the  same  titra- 
tions. 

*  A  1  per  cent  solution  in  alcohol. 

t  Complete  understanding  of  this  subject  involves  the  matter  of  ionic  equilibrium. 
Under  that  head  further  study  will  be  undertaken. 


T1TRATION  OF  POLYBASIC  ACIDS 


57 


Indicators. 

Colors  Exhibited. 

End-point  Concen- 
trations 

In  Acid. 

End  Point. 

in  Alkali. 

(H+). 

(OH-). 

Dimethylamino-azo  ben- 

decided 

salmon 

yellow 

4.9X10"4 

2X10"11 

zene 

pink 

Methyl  orange 

pink 

salmon 

yellow 

2.1X10"4 

4.8X10"1 

Cochineal 

orange 

pink 

lilac 

1.7X10~4 

6X10-11 

Sodium     alizarin     sul- 

brass 

brown 

cherry 

5.2X10-5 

1.9X1Q-10 

phonate 

yellow 

red 

Congo  red 

blue 

violet 

red 

5.2X10"5 

1.9X10-10 

Alizarin 

brass 

old  rose 

pink 

3.7X10-5 

2.7X10~10 

yellow 

Methyl  red 

violet  red 

pink 

yellow 

1.2X10-5 

8.5X10~10 

Azolitmin  (litmus) 

red 

violet 

blue 

10~6 

10~8 

Neutral  red 

magenta 

pink 

yellow 

10~7 

10~7 

Rosolic  acid 

yellow 

rose 

pink 

1.4X10-8 

7X10~7 

Phenolphthalein 

colorless 

pink 

magenta 

9.2X10-9 

1.  1X10~6 

Thymolphthalein 

colorless 

sky  blue 

blue 

5.9X1Q-10 

1.7X10-5 

Trin  itrobenzene 

colorless 

yellow 

orange 

10-i3 

10-1 

Exp.  35.  Titration  of  Polybasic  Acids. 

Procedure  A.  Phosphoric  Acid. — Take  10  cc.  of  approximately  molar 
phosphoric  acid,  and  titrate  with  N  sodium  hydroxide,  using  methyl 
orange  as  indicator.  What  does  the  basicity  of  phosphoric  acid  seem 
to  be  with  this  indicator?  Explain. 

Repeat  the  titration  with  another  10  cc.,  using  phenolphthalein  as 
indicator.  Show  why  phosphoric  acid  appears  to  be  dibasic  in  this  case. 
Since  the  secondary  hydrogen  ion  does  not  redden  methyl  orange,  its 
concentration  must  be  below  a  certain  value.  What  value?  Since 
phenolphthalein  remains  colorless  in  presence  of  this  same  secondary 
hydrogen  ion  the  concentration  of  this  ion  must  be  above  a  certain  value. 
What  value? 

Titrate  a  third  10-cc.  portion  of  the  phosphoric  acid  with  the  alkali, 
using  trinitrobenzene  as  indicator.  Explain  the  result. 

Procedure  B.  Carbonic  Acid. — Carbonic  acid -is  an  example  of  a  very 
weak  dibasic  acid.  Its  primary  ionization  is  too  weak  for  methyl  orange 
but  strong  enough  for  phenolphthalein.  Its  secondary  ionization  is  too 
weak  even  for  the  latter  indicator.  For  the  primary  ionization  we  have 

H2C03-»H++HCO3- 

When  this  primary  hydrogen  is  neutralized,  phenolphthalein  shows  the 
pink  color  with  the  next  drop  of  NaOH,  and  we  then  have  in  solution  the 


58  INDICATORS 

bicarbonate  ion,  HC03~,  such  as  would  be  found  in  a  solution  of  sodium 
bicarbonate,  NaHCO3.  This  may  be  seen  by  preparing  a  clear  solution 
of  sodium  bicarbonate,  and  adding  to  it  a  drop  of  phenolphthalein  indi- 
cator. The  solution  will  be  found  either  to  give  the  pink  color  of  the 
end  point,  or  to  be  ready  to  do  so  upon  the  addition  of  a  drop  of  normal 
alkali. 

A  solution  of  sodium  carbonate  (Na2CO3)  is  alkaline  to  both  methyl 
orange  and  phenolphthalein.  If  we  titrate  this  with  an  acid,  we  have 
the  changes  mentioned  above  occurring  in  reverse  order.  We  have  the 
CO3=  gradually  combining  with  the  H+  ion  from  the  acid  to  form 
HC03-,  thus: 

C03=+H+-^HC03- 

When  this  change  is  completed  the  color  due  to  the  phenolphthalein 
vanishes.  Notice  that  this  end  point  comes  when  the  two  equivalents 
of  carbonate  have  reacted  with  one  equivalent  of  acid.  If  we  now  add 
methyl  orange  and  continue  the  titration,  the  pink  color  of  this  indicator 
will  not  appear  until  the  other  equivalent  is  titrated;  for,  as  the  acid 
is  added,  the  bicarbonate  ion  is  first  changed  to  the  very  slightly  ionized 
carbonic  acid,  which  does  not  affect  the  indicator,  thus : 

HC03-+H+  -»  H2C03 

With  phenolphthalein,  therefore,  sodium  carbonate  acts  like  a  uni- 
valent  or  mon-acid  base,  while  with  methyl  orange  it  acts  like  a  bivalent 
or  di-acid  base.  Since  this  salt  can  easily  be  obtained  pure  and  dry  it  is 
much  used  as  a  standard  in  making  up  standard  acid  solutions. 

Weigh  out  1  gm.  of  pure,  anhydrous  sodium  carbonate,  Na2CO3, 
dissolve  in  about  100  cc.  of  cold  water,  add  phenolphthalein,  and  titrate 
slowly  (note  1)  with  N  hydrochloric  acid.  The  end  point  will  be  a  little 
obscure,  but  if  the  titration  is  carried  out  over  a,  white  tile  and  is  pushed  a 
little  to  the  extreme,  the  result  will  be  exact.  When  this  end  point  is 
reached,  add  methyl  orange  and  proceed  to  the  final  end  point.  Record 
the  amount  of  acid  used  from  the  beginning  up  to  each  end  point,  that  is, 
from  0  to  end  1,  and  from  0  to  end  2.  How  are  these  two  amounts 
related? 

NOTE. — (1)  The  water  should  be  cold  to  prevent  the  escape  of  CO2.  Also,  if 
the  titration  is  too  rapid  the  reaction  may  go  through  both  steps  locally  and  CO2 
may  escape  instead  of  forming  HCCh~~,  as  above  indicated. 


CHAPTER  XII 
HOMOGENEOUS  EQUILIBRIUM 

Exp.  36.  Speed  of  Reaction  and  Speed  Constant. 

We  shall  catalyze  the  hydrolysis  of  methyl  acetate  by  means  of 
hydrochloric  acid,  and  shall  try  to  show  that  the  speed  of  the  reaction  is 
proportional  to  the  concentrations  of  both  the  methyl  acetate  and  the 
hydrogen  ion  We  shall  also  calculate  the  speed  constant,  KI,  for  the 
forward  half  of  the  reaction 

CH3COOCH3  +  H2O+H+  <=>  CH3COOH+CH3OH+H+ 

The  equation  makes  it  appear  as  if  the  speed  of  the  reaction  should 
also  be  proportional  to  the  concentration  of  the  water  present,  and  it  is; 
but  we  shall  arrange  to  have  the  concentration  of  the  water  practically 
the  same  in  all  our  trials,  thus  canceling  out  this  factor  so  far  as  relative 
speed  is  concerned. 

Procedure. — You  have  on  hand  approximately  normal  solutions  of 
NaOH  and  HC1,  which,  if  made  according  to  directions  (Exp.  23), 
correspond  exactly  cubic  centimeter  per  cubic  centimeter  when  methyl 
orange  is  used  as  indicator.  But,  on  account  of  the  presence  of  carbonate, 
this  may  not  be  the  case  if  phenolphthalein  is  used.  Therefore,  since 
phenolphthalein  must  be  used  later  in  titrating  the  acetic  acid  formed 
in  the  reaction,  you  must  know  the  relative  strength  of  the  acid  and 
base  as  shown  by  this  indicator.  To  do  this,  measure  out  10  cc.  of  the 
acid  and  titrate  with  the  NaOH,  using  a  drop  of  phenolphthalein  indi- 
cator. Be  very  exact  about  this. 

Fit  four  150-cc.  flasks  with  good  corks,  and  then  charge  according 
to  the  following  table.  For  measuring  the  acid  use  a  10-cc.  pipette, 
and  for  the  methyl  acetate  a  2-cc.  (dry)  pipette.  Add  the  water  to 
all  the  flasks  first,  then  the  acid,  and  finally  the  acetate.  Proceed 
with  the  last  rapidly,  but  take  care  not  to  let  the  end  of  the  pipette 
enter  the  mixture  so  as  to  transfer  this  to  the  stock  of  methyl 
acetate. 

59 


60 


HOMOGENEOUS  EQUILIBRIUM 


Flask  No. 

Cc.  Water. 

Cc.  N  HC1. 

Cc.  Me  ac. 

1 

48 

10 

2 

2 

38 

20 

2 

3 

46 

10 

4 

4 

36 

20 

4 

As  soon  as  the  ingredients  are  added,  stopper  the  flasks  tightly  with 
wet  corks  (note  1),  and  shake  thoroughly  to  mix  the  contents.  Let  the 
reactions  proceed  for  just  two  hours. 

At  the  end  of  this  time  add  quickly,  in  1-4  order,  sodium  hydroxide 
in  amounts  equal  to  the  acid  originally  used.  This  stops  the  reaction. 
Now  immediately  titrate  the  acetic  acid  formed  in  the  hydrolysis,  using 
phenolphthalein  as  indicator.  The  end  point  will  not  be  permanent, 
because  an  excess  of  methyl  acetate  will  be  present.  Take  as  the  end 
of  the  titration  the  point  where  a  faint  pink  color  remains  for  10  seconds 
after  thorough  stirring.  The  number  of  cubic  centimeters  of  NaOH 
used  in  each  case  may  be  regarded  as  cubic  centimeters  of  acetic  acid, 
and  is,  therefore,  a  measure  of  the  speed  of  reaction  (note  2) . 

It  will  be  noted  that  the  total  volume  in  each  flask  is  60  cc.  There- 
fore, in  flasks  1  and  3,  the  acid  becomes  N/6  and  in  2  and  4  it  becomes 
N/3  when  mixed  with  the  water.  N/6  HC1  is  91  per  cent  ionized,  and 
N/3  is  88  per  cent.  Calculate  the  hydrogen  ion  concentrations,  (H+), 
in  each. 

The  concentrations  of  methyl  acetate  may  be  stated  simply  in  terms 
of  the  cubic  centimeters  used. 

Now  post  the  data  in  tabular  form  as  below,  and  work  out  the  values 
indicated : 


Flask  No. 

(H+). 

(CH3COOCH3) 

Relative 
Speed  Cal- 
culated. 

Cc.  NaOH 
(=  speed) 

Relative 
Speed 
Observed. 

Ki. 

1 

2 

2 

2 

3 

4 

4 

4 

As  indicated  in  the  development  of  the  equilibrium  law,  the  relative 
speeds  may  be  calculated  on  the  assumption  that  they  are  proportional 
to  the  concentrations  of  one  or  both  of  the  reacting  substances.  That  is, 
we  may  let  the  relative  speed  be  1  in  the  first  case;  and  if  in  the  second 


EQUILIBRIUM  CONSTANT  61 

case  the  H+  concentration  is  1.93  times  as  great,  while  that  of  the  acetate 
remains  constant,  we  may  assume  that  the  relative  speed  will  here  be 
1.93,  etc. 

The  observed  relative  speeds  may  be  calculated  from  the  amounts  of 
NaOH  used,  letting  the  first  amount  be  1,  as  above. 

KI  shows  the  relation  between  the  concentration  product  of  the 
reacting  substances  and  the  speed.  It  is  the  same  factor  as  that  sim- 
ilarly represented  in  the  general  equation,  (A)X(B)X  K\  =Si,  and  is 
worked  out  in  the  same  way.  (CH3COOCH3)  goes  in  place  of  (A), 
(H+)  in  place  of  (5),  and  cubic  centimeters  of  NaOH  in  place  of  Si. 
The  value  worked  out  from  these  units  will  probably  be  about  11  for  this 
reaction,  depending  on  the  temperature. 

NOTES. — (1)  Wet  corks  will  not  absorb  the  methyl  acetate. 

(2)  Speeds  are  usually  stated  in  moles  per  minute  or  hour;  here  we  are  stating 
them  in  cubic  centimeters  of  acid  formed  in  two  hours.  Values  obtained  by  different 
people  for  the  same  case  will  probably  differ  somewhat  because  the  temperatures 
used  will  probably  not  be  quite  the  same.  This  does  not  count  against  the  accuracy 
of  any  one  set  of  values  obtained  by  a  single  person,  for  all  the  flasks  used  should 
certainly  be  at  the  same  temperature. 

Exp.  37.  Equilibrium  Constant. 

Note  that  in  36  we  were  not  determining  the  equilibrium  constant 
K,  but  simply  the  speed  constant  K\  for  the  forward  reaction 
CH3COOCH3+H2O+H+^CH3COOH+CH3OH+H+.  To  obtain 
K  we  should  first  need  to  determine  K2  for  the  reverse  reaction.  K 
would  be  the  ratio  between  the  two  constants;  thus  K=  K<z/  K\. 

A  simpler  way  to  determine  the  equilibrium  constant  for  the  hydrolysis 
of  methyl  acetate  would  be  to  allow  the  reaction  to  proceed  until  equi- 
librium had  been  established ;  that  is,  until  the  speeds  of  the  forward  and 
reverse  actions  were  equal.  We  should  then  have  the  equilibrium  rela- 
tions indicated  by  the  equation 

(Meac)X(H20)X(H+) 
(MeOH)X(Hac)X(H+)~ 

It  is  evident  that,  since  all  these  substances  exist  in  the  same  solu- 
tion, the  factor  (H+)  will  be  the  same  for  both  numerator  and  denom- 
inator, and  will,  therefore,  cancel  out.  We  should  then  have 

(Meac)X(H20) 

=  K. 


(MeOH)X(H  ac) 

The  value  for  K  would  be  found,  then,  by  simply  determining  the 
concentrations  of  the  several  factors  which  are  in  equilibrium  as  repre- 
sented in  the  equation,  and  then  solving  in  the  usual  way. 


62  HOMOGENEOUS  EQUILIBRIUM 

We  shall  carry  out  the  experiment  in  two  ways:  first,  by  mixing 
known  amounts  of  the  constituents  represented  in  the  numerator  of  the 
equilibrium  equation,  and  allowing  the  reaction  to  proceed  until  equi- 
librium is  established;  and  second,  by  mixing  known  amounts  of 
the  constituents  represented  in  the  denominator,  and  allowing  equi- 
librium to  be  reached  from  this  side.  The  concentrations  should  so 
adjust  themselves  that  the  value  for  K  will  be  the  same  in  the  two 
cases. 

Procedure. — Take  two  dry  flasks  of  150-cc.  capacity,  fitted  with  tight 
corks,  and  number  them  1  and  2. 

In  flask  No.  1  place  10  cc.  of  methyl  acetate  and  2  cc.  of  water,  each 
measured  accurately  with  pipettes. 

In  flask  No.  2  place  10  cc.  of  glacial  acetic  acid  and  5  cc.  of  methyl 
alcohol. 

Place  in  each  flask  as  a  catalyzer,  1  cc.  of  N  HC1.  This  virtually 
adds  to  each  flask  1  cc.  of  water  which  must  be  allowed  for  later  in  the 
calculation. 

Stopper  the  flasks  loosely,  and  clamp  them  on  a  ring  stand  in  such 
a  position  that  the  bottoms  only  are  immersed  in  water  contained  in  a 
large  graniteware  bath.  Heat  the  water  in  the  bath  to  50°,  and  main- 
tain it  thus  for  three  hours.  This  heating  simply  hastens  the  reactions 
towards  their  equilibrium  points.  When  the  heating  is  well  begun  and 
no  further  pressure  is  likely  to  be  developed,  the  corks  should  be  inserted 
more  tightly  to  prevent  leakage.  After  thus  heating,  let  the  flasks 
stand  where  they  are  until  the  next  afternoon. 

At  the  end  of  this  time,  add  to  each  flask  1  cc.  of  normal  NaOH  to 
neutralize  the  HC1  previously  added,  and  then  proceed  to  titrate  the 
acetic  acid  as  in  Exp.  36,  using  phenolphthalein  as  indicator  and 
taking  the  proper  precautions  about  the  end  point. 

Having  titrated  the  acetic  acid  and  recorded  the  values,  proceed  with 
the  calculations  as  follows: 

Flask  No.  1 

Remembering  that  1  liter  of  normal  alkali  represents  1  mole,  cal- 
culate what  fraction  of  a  mole  of  acetic  acid  was  present.  This  repre- 
sents the  factor  (H  ac)  in  the  equation  above.  The  fraction  of  a  mole  of 
alcohol  formed  will  be  the  same  as  this,  representing  the  factor  (MeOH) . 
To  obtain  the  value  for  the  unhydrolyzed  acetate,  first  find  the  weight 
used  by  multiplying  its  volume  (10  cc.)  into  its  density  (0.964).  From 
this  calculate  the  mole  fraction  used.  The  mole  fraction  still  left 
(Me  ac)  equals  this  minus  the  mole  fraction  of  acetic  acid  or  alcohol 
formed.  To  get  the  factor  (H20)  find  the  initial  mole  fraction  of 


IONIC  EQUILIBRIUM  OF  CUPRIC  BROMIDE  63 

water  present,  and  subtract  from  this  the  mole  fraction  of  acetic  acid 
formed.    . 

Substitute  these  values  in  the  equilibrium  equation,  and  solve  for  K. 

Flask  No.  2 

The  titration  value  for  acetic  acid  converted  into  the  fraction  of  a 
mole  represents  the  factor  (H  ac).  The  initial  molar  concentration  of 
acetic  acid  is  found  by  first  getting  the  weight  of  acid  used  (the  density 
is  1.055)  and  then  converting  to  a  mole  fraction.  The  initial  molar 
concentration  minus  the  factor  (H  ac)  represents  the  fraction  of  a  mole 
transformed  into  acetate,  and  therefore  stands  for  the  factor  (Me  ac). 
The  initial  concentration  of  the  alcohol  is  found  in  the  same  way  (den- 
sity 0.796),  and  this  minus  the  factor  (Me  ac)  represents  the  present 
concentration,  or  mole  fraction,  (MeOH) .  The  factor  (H^O)  will  be  the 
initial  concentration  plus  the  factor  (Me  ac) . 

These  values  are  to  be  substituted  in  the  same  equation  as  those 
from  flask  No.  1,  and  the  value  for  K  is  to  be  worked  out  in  the  same 
way. 

Note  the  significant  fact  that,  despite  the  wide  difference  in  the  con- 
centrations of  the  reacting  substances,  the  ratio  K  really  has  a  nearly 
constant  value. 

Exp.  38.  Ionic  Equilibrium  of  Cupric  Bromide. 

The  following  experiment  is  intended  to  illustrate  the  general  methods 
of  displacing  ionic  equilibrium. 

Procedure. — Weigh  out  on  the  laboratory  balance  6  gm.  of  cupric 
bromide,  CuBr2.  Note  the  color  of  this  substance  in  the  solid  state. 
Place  in  a  small  beaker  and  dissolve  in  15  cc.  of  water.  Cupric  ions, 
Cu++  are  blue  in  color,  bromide  ions,  Br~,  are  colorless,  and  the  non- 
ionized  cupric  bromide  is  black.  To  what  extent  would  you  say,  then, 
that  this  salt  is  ionized  as  you  now  have  it  in  solution?  Write  the 
equilibrium  equation  for  the  ionization  of  cupric  bromide. 

Take  5  cc.  of  the  concentrated  solution,  add  water  in  2-cc.  portions, 
shaking  after  each  addition,  until  10  cc.  have  finally  been  added.  Note 
and  explain  the  successive  changes  in  color. 

Take  another  5-cc.  portion  of  the  concentrated  solution,  and  add  an 
equal  volume  of  water,  obtaining  an  emerald-green  solution.  To  this 
add  about  2  gm.  of  sodium  bromide.  Explain  the  change  in  color. 
What  is  the  general  term  applied  to  this  method  of  displacing  ionic 
equilibrium? 

To  the  final  portion  of  the  concentrated  solution  add  2  gm.  of  pow- 


64  HOMOGENEOUS  EQUILIBRIUM 

dered  cadmium  nitrate,  and  stir  until  dissolved.  We  shall  then  have  the 
following  equilibria: 

CuBr2<=±Cu+++2Br- 

Cd(N03)2  <=±2NO3-+Cd+  + 

IT  IT 

Cu(NO3)2CdBr2 

Cupric  nitrate  is  highly  ionized,  so  we  shall  not  look  for  any  disturbance 
of  the  equilibrium  from  that  source.  But  cadmium  bromide  is  only 
weakly  ionized;  therefore  a  considerable  amount  of  the  non-ionized  salt 
will  at  once  be  formed,  thus  lowering  the  concentration  of  the  bromide 
ion.  What  effect  does  this  have  upon  the  upper  equilibrium?  What 
effect  on  the  concentration  of  the  cupric  ion,  Cu++,  and  the  non-ionized 
cupric  bromide,  CuBr2?  What  evidence  do  you  have  that  your  expla- 
nations are  correct?  What  name  is  applied  to  this  method  of  displacing 
ionic  equilibrium? 

Exp.  39.  Common  Ion  Effect  with  Acetic  Acid  and  Ammonium  Hydrox- 
ide. 

Procedure  A. — From  the  ionization  table  (text)  find  the  degree  of 
ionization  of  molar  acetic  acid.  From  this  calculate  the  concentrations 
of  the  ions  and  the  non-ionized  part,  and  then  work  out  the  ionization 
constant.  Note  that  the  ion  product  is  practically  equal  to  the  ioniza- 
tion constant  in  this  case.  Would  this  be  true  in  any  other  than  a  molar 
solution? 

Measure  out  50  cc.  of  molar  acetic  acid,  and  add  to  it  enough  sodium 
acetate  (NaC2H302-3H20)  to  make  the  concentration  of  this  salt  0.2 
molar  (M/5),  stirring  until  dissolved.  Again  referring  to  the  ionization 
table,  calculate  the  acetate  ion  concentration  coming  from  the  acetate, 
and  then  the  total  acetate  ion  concentration.  Having  done  this,  cal- 
culate the  quantitative  effect  this  increase  of  acetate  ion  must  have  on 
the  hydrogen  ion  concentration  which  must  still  be  in  equilibrium  with  it. 
What  color  should  methyl  orange  show  in  this  solution?  Test  your  cal- 
culation by  adding  a  drop  of  the  indicator  and  stirring. 

What  would  be  the  effect  of  adding  still  more  sodium  acetate? 
What  concentration  of  hydrogen  ion  would  have  been  present  in  the 
above  solution  if  enough  sodium  acetate  had  been  added  to  make  the 
concentration  of  this  salt  molar?  (Molar  sodium  acetate  is  53  per  cent 
ionized.) 

Procedure  B. — Let  us  assume  that  our  problem  here  is  to  calculate 
how  much  ammonium  chloride  must  be  added  to  50  cc.  of  N/10  NHUOH 
to  reduce  the  OH~  ion  concentration  to  the  turning  point  for  phenol- 


NEUTRALIZATION  65 

phthalein.  (1.1X10~6.)  Substitute  values  for  N/10  ammonium  hy- 
droxide in  the  equilibrium  equation 

(NH4+)X(OH-) 
(NH4OH) 

and  then  calculate  the  value  of  the  product 

(NH4+)X(OH-) 

Assuming  that  this  product  remains  practically  constant  (prove  this), 
calculate  what  total  concentration  of  NH4+  ion  will  be  necessary  to 
reduce  that  of  the  OH~  to  the  desired  value.  You  will  find  this  con- 
centration to  be  something  like  1.7.  To  get  this  concentration  of  NH4+ 
you  will  need  to  use  ammonium  chloride  of  such  a  concentration  that 
it  will  probably  not  be  more  than  70  per  cent  ionized  (note  1).  On  the 
assumption  that  the  salt  is  70  per  cent  ionized,  calculate  the  total  con- 
centration necessary  to  give  the  proper  NH4+  concentration.  Having 
done  this,  you  can  at  once  determine  the  number  of  grams  of  NH4C1 
necessary  to  give  this  concentration  in  50  cc. 

When  you  are  sure  that  your  value  is  correct,  weigh  out  the  proper 
amount  of  ammonium  chloride,  place  in  a  small  beaker,  and  dissolve 
in  50  cc.  of  N/10  ammonium  hydroxide.  Finally  add  a  drop  of  the 
indicator.  The  color  should  be  a  barely  perceptible  pink — the  end- 
point  color. 

NOTE. — 3  N  ammonium  chloride  is  68  per  cent  ionized,  and  therefore  gives  an 
NH,+  concentration  of  2.04. 

Exp.  40.  Neutralization  by  Formation  of  a  Non-ionized  Acid  or  Base. 
Procedure  A. — Measure  out  50  cc.  of  normal  HC1  into  a  beaker. 
What  concentration  of  hydrogen  ion  is  present  in  this  solution?  Add  a 
drop  of  methyl  orange  and  note  the  intense  red  color  shown  by  the  indi- 
cator. Now  add  to  the  solution  sufficient  sodium  acetate  to  make  the 
solution  1.8  molar  with  reference  to  this  salt  and  stir  until  all  is  in  solu- 
tion. Note  the  effect  on  the  indicator.  What  has  become  of  the 
hydrogen  ion?  Suppose  we  consider  the  reaction  as  taking  place  in  two 
steps:  first,  1  mole  of  acetate  reacting  with  1  mole  of  acid;  and  second, 
the  other  0.8  mole  of  acetate  reacting.  At  the  end  of  the  first  step ,  what 
concentration  of  hydrogen  ion  will  be  present?  To  what  extent  will  the 
acid  have  been  neutralized?  Calculate  the  quantitative  effect  of 
the  other  0.8  mole  of  acetate  on  the  hydrogen  ion  concentration  by  the 
method  employed  in  Exp.  39,  counting  the  salt  as  40  per  cent  ionized. 
What  concentration  of  hydrogen  ion  should  be  present?  Does  this 


66  HOMOGENEOUS  EQUILIBRIUM 

account  for  the  color  shown  by  the  indicator?  What  indicator  would 
practically  show  its  end-point  color  in  this  solution?  Test  your  con- 
clusion if  the  indicator  is  available.  The  test  may  be  made  in  presence 
of  the  methyl  orange. 

Procedure  B. — This  procedure  should  refer  to  the  neutralization  of 
some  strong  base,  say  sodium  hydroxide,  by  the  addition  of  the  salt  of  a 
weak  base.  This  problem  is  left  for  the  student  to  work  out,  with  the 
suggestion  that  N/10  sodium  hydroxide  be  used  to  start  with,  and  that 
it  be  the  problem  to  calculate  what  weight  of  NEUCl  must  be  added  to 
50  cc.  to  reduce  the  OH~  concentration  to  the  turning  point  for  thymol- 
phthalein  (1.7X10~5).  The  calculation  is  best  made  in  two  steps,  as 
was  done  with  acetic  acid  in  Procedure  A.  This  should  afterwards  be 
tested  by  experiment,  of  course.  In  order  to  see  the  very  faint  blue 
color  of  the  indicator  at  its  turning  point,  it  may  be  necessary  to  com- 
pare the  solution  side  by  side  with  pure  water. 

Exp.  41.  Hydrolysis  of  Salts. 

Procedure  A.  Qualitative. — Test  solutions  of  the  following  salts  with 
red  and  blue  litmus  solution,  and  note  whether  the  reaction  is  neutral, 
acid,  or  alkaline:  sodium  acetate,  potassium  carbonate,  aluminum 
chloride,  sodium  chloride,  ammonium  chloride. 

It  may  be  necessary  in  some  cases  to  compare  the  solution  containing 
the  indicator  with  pure  water  containing  the  indicator  alone.  If  this 
is  necessary  place  the  two  solutions  in  clean  test-tubes  and  look  down 
through  them  as  you  hold  them  side  by  side  over  a  piece  of  white  paper. 

Write  the  proper  equilibria,  and  explain  each  case  carefully.  Make 
some  estimate  also  as  to  the  relative  degrees  of  hydrolysis  in  the  different 
cases. 

Procedure  B.  The  Degree  of  Hydrolysis  of  Aniline  Sulphate.— 
Aniline  sulphate  is  the  salt  of  a  weak  base,  aniline  hydroxide,  and  the 
strong  acid,  sulphuric.  Its  solution  will,  therefore,  show  an  accumula- 
tion of  H+  ion.  Our  problem  will  be  to  measure  the  concentration  of 
this  H+  ion,  and  from  this  to  calculate  the  degree  of  hydrolysis. 

The  color  comparator  and  six  Nessler's  tubes  will  be  needed. 

Prepare  1  liter  of  N/1000  hydrochloric  acid,  and  then  make  up  five 
standard  tubes  containing,  respectively,  100  cc.,  80  cc.,  60  cc.,  40  cc., 
and  20  cc.,  of  this  acid,  filling  each  to  the  100  cc.  mark  with  distilled 
water.  Prepare  also  100  cc.  of  N/10  aniline  sulphate  solution  (equiva- 
lent weight  142),  and  place  in  the  sixth  Nessler's  tube.  The  temperature 
of  this  solution  should  be  approximately  25°  C.  The  tube  containing  it 
should  be  labeled  to  prevent  confusing  with  the  others.  Arrange  the 
tubes  in  the  comparator  so  that  the  alternate  holes  will  be  left  vacant. 


HYDROLYSIS  OF  SALTS  67 

Now  add  to  each  tube  four  drops  of  methyl  orange  indicator,  and 
mix  in  the  usual  way.     Having  done  this,  match  colors  between  the 
aniline  sulphate  solution  and  the  acid  solutions,  and  from  the  data  thus 
obtained  determine,  as  accurately  as  possible,  the  hydrogen  ion  concen 
tration  resulting  from  hydrolysis. 

The  calculation  of  the  degree  of  hydrolysis  may  be  made  as  follows: 
(a)  First  calculate  the  theoretical  degree  from  the  known  constants. 
It  has  been  shown  in  the  classroom  discussion  of  hydrolysis  that  in  a  case 
like  this  the  following  relation  holds  : 

(total  salt-x)<*i  Kbase 


(strong  acid)  <*2  (weak  base)       Kw 

where  a\  and  0:2  are  the  degrees  of  ionization  of  the  salt  and  strong  acid 
respectively,  where  Xbase  is  the  ionization  constant  for  the  weak  base, 
and  x  is  the  concentration  of  strong  acid  or  weak  base  formed.  It  is 
evident  without  discussion  that  the  left-hand  member  of  this  equation 
may  go  into  the  form 

(total  salt  —  x)ai 

X2a2  ' 

It  has  also  been  shown  that  where  the  value  of  K^ase  is  not  smaller 
than  10~10  the  degree  of  hydrolysis  will  be  small,  and  that  in  such  a  case 
the  concentration  (total  salt  —  x)  may  be  taken  as  equal  to  (total  salt). 
Such  is  the  case  with  aniline  sulphate,  for  K^e  here  is5XlO~10.  We 
shall,  therefore,  write  the  equation 

(total  salt)ai 


The  necessary  constants  are  as  follows  : 

(Total  salt)  =         0.1      (Normal  cone.)  . 

ai  0.7      (Ionization  of  N/10  An2SO4). 

a2  0.61    (Ionization  of  N/10  H2SO4)  (note  1). 

^base  =     5X10-10 

Kw  10-14 

Substitute  these  values  in  equation  (2)  and  work  out  the  value  of  x. 
This  will  be  the  concentration  (weak  base)  or  (strong  acid).  When  it  is 
remembered  that  each  equivalent  of  salt  hydrolyzed  gives  one  equivalent 
of  acid  or  base,  the  per  cent  hydrolyzed,  or  degree  of  hydrolysis,  may  be 
calculated  at  once. 

(6)  Note  now  that  the  H+  ion  concentration,  as  shown  in  (a),  is 
0.61  of  the  total  concentration  of  acid  (or  base)  formed.  From  the 


68  HOMOGENEOUS  EQUILIBRIUM 

hydrogen  ion  concentration  which  you  have  just  determined  experi- 
mentally, you  may  now  calculate  the  total  concentration  of  acid  or  base 
formed  and  then  the  per  cent,  or  degree,  of  hydrolysis.  The  value  thus 
obtained  should  agree  well  with  the  value  calculated  in  (a) . 

NOTE. — (1)  The  concentration  of  the  sulphuric  acid  formed  is  not  N/10;  but  in 
the  presence  of  the  large  concentration  of  the  SO4=  ion  from  the  salt,  the  degree  may 
be  considered  the  same  as  for  an  N/10  solution. 


CHAPTER  XIII 
HETEROGENEOUS  EQUILIBRIUM 

Exp.  42.  Decomposition  of  the  Hydrates  of  Cupric  Sulphate. 

Select  clear,  blue  crystals  of  the  pentahydrate  of  cupric  sulphate 
showing  no  efflorescence.  Crush  in  a  mortar,  and  weigh  out  a  sample  of 
1  gm.  into  a  crucible  which  has  been  previously  weighed  with  its  cover. 
Heat  in  an  air  bath  at  a  temperature  of  130°  C.  for  forty-five  minutes, 
cool  in  a  desiccator,  and  weigh  with  cover  in  place.  Repeat  the  heating  to 
constant  weight.  Assuming  that  the  original  salt  was  the  pentahydrate, 
calculate  the  number  of  molecules  of  water  lost  at  130°  C.  The  calcu- 
lation is  as  follows : 

Let  x  =  number  of  molecules  of  H2O  lost.  The  equation  showing  the 
decomposition  will  then  be 

CuSO4-5H2O-o;H20  -»  CuS04  aq. 
and 

249.7  -18z^l59.6+aq. 

From  this  we  obtain 

249.7  :  18x::l  :  wt.  of  water  lost:  from  which  x.  the  number  of 
molecules,  may  be  calculated. 

Why  is  it  necessary  to  heat  the  pentahydrate  to  make  it  give  off 
water?  How  could  the  same  result  be  obtained  without  heating? 
What  equilibrium  conditions  must  be  met  in  order  to  cause  the  hydrate 
left  from  the  above  reaction  to  lose  its  water?  How  can  this  be  done? 
Would  it  require  some  definite  temperature  (assuming  that  the  humidity 
of  the  air  remains  constant)  or  would  any  slight  increase  above  130°  C. 
bring  it  about?  Why?  What  if  the  humidity  of  the  air  became  much 
less? 

Test  your  conclusion  by  heating  the  hydrate  left  in  the  crucible  to 
200°  C.  for  fifteen  minutes.  (The  humidity  of  the  air  is  not  likely  to 
have  changed  if  this  is  done  the  same  day.)  After  cooling  in  a  desic- 
cator, weigh.  Explain  the  result. 

Now  heat  to  250°  for  thirty  minutes,  cool  and  weigh.  Finally  heat  to 
constant  weight  at  this  temperature.  Now  again  substitute  values  in 
the  equation  above  and  calculate  the  total  number  of  molecules  of  water 
lost  at  250°  C. 


70  HETEROGENEOUS  EQUILIBRIUM 

Exp.  43.  Partition  of  Bromine  between  Water  and  Carbon  Tetrachloride. 

Apparatus. — Two  100-cc.  glass-stoppered  bottles.  (Obtain  from 
instructor.) 

Procedure. — We  shall  take  a  solution  of  bromine  in  water  and  first 
determine  its  concentration.  We  shall  then  bring  this  solution  in  con- 
tact with  carbon  tetrachloride  in  two  different  proportions.  The  bro- 
mine will  distribute  itself  between  these  two  solvents  according  to  the 
equilibrium  law,  which  demands  that  the  ratio  between  the  concen- 
trations in  the  two  solvents  shall  be  constant,  although  these  concen- 
trations may  differ  widely  in  the  two  cases. 

Prepare  a  nearly  saturated  solution  of  bromine  in  water,  or  use  the 
reagent  "  bromine  water."  Prepare  also  an  approximately  N/10  solu- 
tion of  sodium  thiosulphate.  (See  Procedure  A,  Exp.  18,  p.  35.) 

To  determine  the  concentration  of  the  bromine  water,  remove  5  cc. 
with  a  pipette  and  transfer  to  a  small  Erlenmeyer  flask  containing  10  cc. 
of  10  per  cent  solution  of  potassium  iodide.  (This  bromine  solution 
should  be  transferred  directly  from  the  stock  bottle  to  the  potassium 
iodide  solution  as  directed.  It  must  not  be  carried  in  an  open  beaker  or 
even  placed  in  the  flask  before  the  potassium  iodide  solution  is  put  in. 
If  it  is,  bromine  will  be  lost.)  The  iodine  set  free  by  the  bromine  is 
titrated  with  the  thiosulphate.  Repeat  the  determination,  and  record 
as  the  concentration  of  the  bromine  water  the  average  number  of 
cubic  centimeters  of  thiosulphate  required.  All  the  concentrations  we 
need  to  determine  in  this  experiment  will  be  stated  in  the  same  way: 
"  cubic  centimeters  of  thiosulphate  equivalent  to  5  cc.  of  solution." 

Now  place  in  one  of  the  glass-stoppered  bottles  5  cc.  of  carbon 
tetrachloride,  measured  very  accurately  with  a  dry  pipette.  In  the 
other  place  10  cc.  of  the  same  liquid.  To  each  bottle  add  50  cc.  of  the 
bromine  water  (pipette),  and  quickly  insert  the  stoppers.  Holding  the 
stoppers  in  place,  shake  the  bottles  vigorously  for  ten  minutes,  and  then 
let  stand  for  half  an  hour  for  the  two  solvents  to  separate  completely. 
At  the  end  of  this  time  remove  a  5-cc.  portion  of  the  water  layer  from 
each  bottle,  and  proceed  as  above  to  determine  the  concentration. 
Record  these  titrations  as  (Brw) .  To  get  the  concentrations  in  the  other 
solvent  proceed  as  follows:  You  have  the  titration  for  5  cc.  of  the  original 
bromine  water.  From  this  calculate  the  total  value  for  50  cc. ;  that  is, 
all  the  bromine  present  in  each  case  as  cubic  centimeters  of  thiosulphate. 
In  the  same  way  calculate  the  value  for  50  cc.  after  partition  in  each  case. 
The  difference  between  the  total  bromine  content  before  partition  and 
after  partition  will  be  the  bromine  dissolved  in  the  carbon  tetrachloride 
layer,  stated,  of  course,  as  "  cubic  centimeters  of  thiosulphate."  Record, 
as  the  concentration  of  the  bromine  in  the  carbon  tetrachloride,  the 


PARTITION  OF  SUCCINIC  ACID  71 

number  of  cubic  centimeters  thiosulphate   equivalent  to  5  cc.  of  this 
layer.     This  is  designated  as  (Brc). 

After  obtaining  the  two  sets  of  values  for  (Brw)  and  (Brc),  substitute 
in  the  equation 

(Brc)/(Brw)  =  K 

and  calculate  the  values  for  K.     They  will  be  nearly  identical  if  the  tem- 
perature was  the  same  in  both  bottles,  and  will  not  differ  widely  from  25. 

Exp.  44.  Partition  of  Succinic  Acid  between  Water  and  Ether. 

Apparatus. — Two  100-cc.  glass-stoppered  bottles,  as  in  Exp.  43. 

Procedure. — Place  in  bottle  No.  1  about  20  cc.  of  a  solution  of 
succinic  acid  of  about  M/10  concentration.  In  bottle  No.  2  place  10  cc. 
of  the  same  solution,  and  then  add  about  an  equal  volume  of  water. 
Now  add  to  each  bottle  20-25  cc.  ether.  Placing  the  stoppers  in  the 
bottles  and  holding  them  securely  in  place,  shake  the  bottles  vigorously 
for  not  less  than  five  minutes  to  bring  about  the  distribution  of  the  suc- 
cinic acid  between  the  two  solvents.  When  this  has  been  done,  let  the 
two  solutions  stand  a  moment  for  the  layers  to  separate,  and  then  pro- 
ceed to  determine  the  concentrations  in  the  several  cases  as  follows : 

By  means  of  a  dry  5-cc.  pipette,  remove  a  sample  (5  cc.)  of  one  of  the 
upper  layers  (ether),  place  in  a  small  Erlenmeyer  flask,  add  about  50  cc. 
distilled  water  and  a  drop  of  phenolphthalein  indicator,  and  then  titrate 
very  carefully  with  approximately  N/10  NaOH  (note). 

Draw  out  5  cc.  of  the  lower  layer  in  the  same  bottle  and  titrate  in 
the  same  way.  When  the  tip  of  the  pipette  is  passing  through  the  upper 
layer,  this  solution  should  be  excluded  by  closing  the  upper  end  with  the 
finger  tip. 

Having  titrated  the  two  layers,  let  the  number  of  cubic  centimeters  of 
alkali  used  represent  the  concentration  in  each  case,  and  then  work  out 
the  partition  ratio,  Cwater/Cether. 

Now  proceed  in  the  same  way  to  determine  the  concentrations  and 
the  partition  ratio  for  the  solutions  in  the  other  bottle. 

The  ratios  in  the  two  cases  should  be  almost  exactly  equal,  although 
difference  in  temperature  may  cause  them  to  differ  slightly  from  those 
found  by  another  worker. 

NOTE. — This  can  be  made  up  by  dilution  of  the  normal  alkali.  It  is  important 
that  the  titration  be  accurate,  even  though  the  alkali  used  be  of  unknown  concen- 
tration: the  volumes  must  be  accurately  known. 

Exp.  45.  The  Cooling  Curve  of  Sodium  Sulphate  in  the  Light  of  the 

Phase  Rule. 

Apparatus. — A  large  test-tube  (25X180  mm.);  a  ring  stand;  two 
common  thermometers;  a  300-cc.  beaker  of  tall  form;  a  small  reading 


72 


HETEROGENEOUS  EQUILIBRIUM 


lens;  a  swab  made  by  winding  a  little  cotton  around  the  end  of  a  glass 
stirring  rod. 

Procedure  A.  (note  1.) — Take  enough  sodium  sulphate  decahydrate 
to  make  a  layer  in  the  test-tube  about  7  cm.  deep.  Fasten  the  apparatus 
up  on  the  ring  stand,  immersing  the  tube  in  water  above  the  top  of 
the  sulphate.  One  thermometer  measures  the  temperature  of  the  water 
and  the  other  that  of  the  salt. 

Heat  the  water  to  45°  C.  and  maintain  this  temperature  until  the 
sulphate  has  apparently  melted  down  and  the  temperature  of  the  inner 
thermometer  reads  nearly  the  same  as  the  outer  one.  Both  the  water 
and  the  salt  should  be  frequently  stirred.  During  this  process  of  heating 
the  sulphate  has  given  off  its  water  of  hydration,  becoming  anhydrous 
and  in  this  water  a  part  of  the  anhydrous  salt  has  dissolved. 

Now  remove  the  beaker  containing  the  water  and  wipe  off  the  outside 
of  the  test-tube.  The  temperature  will  immediately  begin  to  fall. 
Read  the  thermometer  every  two  minutes  until  it  has  fallen  to  about 
32.5°,  using  the  lens  and  wiping  off  the  inside  of  the  tube  with  the  swab  if 
necessary.  The  transition  from  the  anhydrous  form  to  the  decahydrate 
occurs  at  32.4°  C.  (note  2),  but  the  transformation  may  be  suspended. 
To  prevent  this,  drop  in  about  1  gm.  (estimated)  of  decahydrate.  From 
this  time  on  take  a  reading  every  ten  minutes,  stirring  well  each  time  and, 
as  far  as  possible,  keeping  the  salt  from  sticking  to  the  bottom  of  the 
tube.  When  the  mass  becomes  very  thick  it  will  be  best  to  stop  stirring 
and  simply  leave  the  thermometer  embedded  in  the  middle  of  the  tube. 
After  the  transition  is  complete  and  the  temperature  again  begins  to 
change,  read  every  five  minutes  until  the  temperature  nearly  reaches 
that  of  the  room. 

All  these  readings  should  be  recorded  in  the  notebook  thus : 


Time. 

Temperature. 

2 

45° 

4 

etc. 

etc. 

30 

40 

50 

etc. 

At  the  end  of  the  experiment  the  values  should  be  plotted  on  coordinate 
paper,  and  a  smooth,  neat  time-temperature  curve  drawn. 


THE  COOLING  CURVE  OF  SODIUM  SULPHATE  73 

Explain  the  shape  of  the  curve  in  the  light  of  the  phase  rule,  as  follows : 

The  system  is  made  up  of  two  components — What  are  they?  At  45° 
C.  there  are  three  phases — What  are  they?  What  is  the  variance  of  a 
system  having  two  components  and  three  phases?  How  does  the  first 
part  of  the  curve  indicate  this?  At  32.4°  C.  another  phase  appears — 
What?  We  then  have  four  phases  and  two  components.  What  is  the 
variance  of  such  a  system?  How  does  the  curve  show  this?  When  the 
transformation  is  complete  one  phase  drops  out — Which  one? — and 
the  system  again  becomes  three-phase.  What  is  the  variance,  and  how 
does  the  curve  show  it? 

Procedure  B. — Proceed  as  in  A,  with  one  exception:  Do  not  stir 
the  mixture  after  removal  of  the  tube  from  the  beaker  of  water.  Be 
careful  about  this,  even  in  reading  the  thermometer.  It  will  be  well,  also, 
to  have  the  mouth  of  the  tube  plugged  with  cotton. 

You  note,  of  course,  that  the  transformation  remains  suspended,  the 
fourth  phase  not  appearing  at  all  (note  3).  The  system  therefore 
remains  three-phase  and  univariant  throughout.  The  data  should  be 
recorded  and  plotted  as  in  A,  when  this  will  at  once  be  seen. 

When  the  temperature  has  dropped  nearly  to  that  of  the  room,  it  will 
be  interesting  to  note  what  will  happen  if  the  fourth  phase  (the  deca- 
hydrate)  is  introduced.  To  this  end,  drop  in  about  a  gram  of  the 
hydrate ;  then  stir  the  mixture  and  proceed  to  read  temperatures  at  ten- 
minute  intervals.  Plot  and  explain  the  data. 

NOTES. — (1)  This  experiment  will  require  a  full  laboratory  period  of  three  hours 
and  a  half. 

(2)  It  may  not  be  quite  this  by  your  thermometer,  because  the  instrument  may 
not  be  very  accurate. 

(3)  The  fourth  phase  may  appear  spontaneously.     If  it  does  it  will  only  be  neces- 
sary to  stir  and  continue  reading  the  temperatures  as  directed. 


CHAPTER  XIV 
COMPLEX  EQUILIBRIUM 

Exp.  46.  Precipitation  and  Solution  of  Silver  Acetate. 

Procedure  A. — Calculate  whether  by  the  mixing  of  equal  volumes  of 
N/5  sodium  acetate  and  N/10  silver  nitrate  a  precipitate  of  silver  acetate 
should  be  formed.  Proceed  as  follows: 

A  saturated  solution  of  silver  acetate  at  16°  C.  is  0.0603  molar,  and 
the  salt  in  this  solution  is  71  per  cent  ionized.  Calculate  the  solubility 
product  (Ag+)  X  (ac~~)  and  the  concentration  of  the  non-ionized  mole- 
cules (Ag  ac). 

When  N/5  sodium  acetate  is  mixed  with  N/10  silver  nitrate  the 
former  becomes  N/10  and  the  latter  N/20.  For  the  ionization  of  these 
salts  at  this  concentration  consult  the  Ionization  Table  in  the  text. 
Determine  the  concentrations  of  the  silver  ion  and  the  acetate  ion 
respectively. 

Now,  if  a  precipitate  is  to  be  formed,  the  ions  must  be  able  to  unite 
in  sufficient  amount  to  more  than  saturate  the  solution  with  non-ionized 
molecules  and  still  leave  the  ion  product  equal  to  the  solubility  product. 
If,  then,  x  is  the  largest  concentration  of  molecules  that  may  be  formed 
and  still  leave  these  conditions  fulfilled,  the  relation  indicated  by  the 
following  equation  must  hold: 

[(Ag+)-*]X[(ac-)-*]-S.P. 

To  tell  whether  a  precipitate  will  be  formed  we  have,  therefore,  only  to 
substitute  values  in  this  equation  and  find  whether  x  exceeds  the  value 
of  (Ag  ac)  for  a  saturated  solution. 

Having  made  the  calculation,  test  the  validity  of  your  conclusion  by 
actually  mixing  the  two  solutions  (20  cc.  of  each),  and  taking  the  proper 
precautions  against  super  saturation. 

Procedure  B. — Calculate  whether  the  mixing  of  20  cc.  of  N/5  sodium 
acetate  with  10  cc.  of  N/5  silver  nitrate  should  yield  a  precipitate  of 
silver  acetate.  Note  that  when  these  solutions  are  mixed,  the  volume, 
so  far  as  the  sodium  acetate  is  concerned,  becomes  3/2  as  great,  making 
the  concentration  2/3  the  original  concentration.  For  the  silver  nitrate 

74 


PRECIPITATION  BY  MEANS  OF  HYDROGEN  SULPHIDE         75 

the  volume  becomes  three  times  as  great,  giving  the  concentration  1/3 
its  original  value.  The  degree  of  ionization  may  be  obtained  from  the 
table  by  interpolation.  Having  done  this,  calculate  the  concentration 
of  the  two  ions,  and  then  proceed  as  in  A  to  determine  whether  a  pre- 
cipitate should  be  formed.  Finally  test  the  calculation  by  actual  ex- 
periment. 

Procedure  C. — If  a  precipitate  is  produced  in  B,  the  mixture  thus 
obtained  may  be  used  here.  Before  going  on,  however,  let  the  mixture 
stand  for  ten  minutes  with  frequent  stirring  to  make  sure  that  no  super- 
saturation  exists.  Proceed  then  as  follows:  Decant  most  of  the  liquid 
through  a  filter,  leaving  the  remainder  with  the  precipitate. 

To  the  portion  containing  the  precipitate,  add  three  drops  of  concen- 
trated nitric  acid,  and  stir.  Explain  the  result. 

The  filtrate  is  a  saturated  solution  of  silver  acetate.  What  effect 
should  it  have  to  add  a  little  solid  sodium  acetate?  Carry  out  the 
experiment,  adding  1  gm.  of  the  salt  (do  not  weigh),  and  stirring  until 
dissolved.  Explain  the  result. 

Exp.  47.  Precipitation  by  Means  of  Hydrogen  Sulphide  and  its  Salts. 

Procedure  A. — Take  30  cc.  of  M/10  zinc  sulphate  solution  in  a  small 
Erlenmeyer  flask  and  pass  H^S  gas  through  it  rapidly  for  five  minutes. 
(This  should  be  done  in  a  hood  where  you  are  sure  there  is  a  good  draft.) 
Filter  off  the  precipitate  of  zinc  sulphide  and  pass  EbS  through  the  fil- 
trate for  five  minutes.  Repeat  the  process  until  nothing  more  can  be 
precipitated  in  this  way. 

Test  the  clear  filtrate  with  litmus,  and  explain  the  presence  of  the 
hydrogen  ion,  writing  the  proper  equilibria. 

Now  divide  the  filtrate  into  two  parts.  To  one  part  add  an  excess  of 
ammonium  hydroxide,  and  to  the  other  part  add  solid  sodium  acetate. 
Was  the  zinc  all  precipitated  by  the  H^S?  Does  the  presence  of  the 
hydrogen  ion  noted  above  have  anything  to  do  with  this?  Explain 
very  carefully,  writing  the  proper  equilibria.  The  precipitate  formed 
when  ammonium  hydroxide  was  added  was  zinc  sulphide,  not  hydroxide. 
Explain  its  formation.  Why  was  the  same  thing  formed  when  sodium 
acetate  was  added? 

Take  another  30  cc.  of  the  zinc  sulphate  solution,  add  an  equal  vol- 
ume of  normal  HC1,  and  pass  H2S  through  it  for  five  minutes.  Explain 
the  result.  Why  is  zinc  not  precipitated  by  H2&  along  with  the  members 
of  the  Copper  Group  in  qualitative  analysis? 

Procedure  B. — Take  30  cc.  of  M/10  cupric  sulphate  solution,  add 
an  equal  volume  of  normal  HC1,  and  pass  H2S  through  it  rapidly  for 
five  minutes.  Filter  off  the  precipitate  of  cupric  sulphide,  and  pass  EbS 


76  COMPLEX  EQUILIBRIUM 

through  the  filtrate  for  five  minutes.  Repeat  the  process  until  satisfied 
that  nothing  more  can  be  precipitated  in  this  way. 

Now  test  the  filtrate  as  follows  to  see  whether  the  copper  has  been 
completely  removed:  Boil  for  about  five  minutes  to  remove  the  K^S, 
and  then  cool  by  rotating  the  flask  under  a  stream  of  cold  water.  Now 
divide  into  two  parts.  To  one  part  add  a  bright  piece  of  sheet  zinc  and 
let  stand  for  a  time.  The  zinc  will  show  a  red  coating  of  metallic  copper 
when  the  concentration  is  as  low  as  0.000008  gm.  per  cubic  centimeter. 
Is  there  any  evidence  of  its  presence  in  this  case?  To  the  other  portion 
add  solid  sodium  acetate  (about  1  gm.).  This  will  nearly  neutralize 
the  free  acid  present.  (Why?)  Now  add  potassium  ferrocyanide 
solution  (1  cc.).  If  copper  is  present  a  cherry-red  color  will  appear. 
This  test  will  show  the  presence  of  0.00001  gm.  of  copper  per  cubic 
centimeter. 

Why  is  copper  precipitated  completely,  even  in  the  presence  of 
hydrochloric  acid,  while  zinc  was  not,  even  when  starting  with  a  neutral 
solution? 

Procedure  C. — Take  30  cc.  of  M/10  ferrous  ammonium  sulphate 
solution  (freshly  prepared),  and  pass  H^S  through  it.  Is  FeS  (black) 
precipitated?  Add  ammonia  to  the  solution,  or  add  ammonium  sul- 
phide to  a  fresh  portion.  Explain  the  result. 

Exp.  48.  Precipitation  of  Magnesium  by  Means  of  Ammonium  Hydrox- 
ide. 

Procedure  A. — Magnesium  hydroxide  is  soluble  in  water  to  the  extent 
of  0.009  gm.  per  liter  at  18°  C.  Calculate  this  into  terms  of  moles  per 
liter  (molar  solubility).  Assuming  that  the  base  is  completely  ionized 
at  this  great  dilution,  calculate  first  the  concentrations  of  the  ions  Mg++ 
and  OH~,  and  from  these  calculate  the  solubility  product  (Mg++)X 
(OH-)2. 

Suppose  we  were  to  mix  molar  ammonium  hydroxide  with  an  equal 
volume  of  M/5  magnesium  sulphate,  the  former  becoming  thereby 
M/2  and  the  latter  M/10,  would  a  precipitate  of  Mg(OH)2  be  formed? 
M/2  NH4OH  is  0.57  per  cent  (0.0057)  ionized.  M/10  MgSO4  is  37 
per  cent  (0.37)  ionized.  Calculate  the  concentrations  of  the  two  ions 
Mg++  and  OH~  resulting  from  this  ionization,  and  then  calculate  the 
ion  product  (Mg++)  X  (OH~)2.  Comparing  this  ion  product  with  the 
solubility  product  calculated  above,  and  making  no  allowance  for  the 
negligible  amount  of  non-ionized  Mg  (OH) 2,  decide  whether  a  precip- 
itate should  be  formed.  Finally  mix  20  cc.  of  each  of  the  original 
solutions  and  note  the  result. 

Is  the  magnesium  all  precipitated  by  the  above  process,  or  is  there 


THE  SILVER  AMMONIUM  COMPLEX  77 

still  a  considerable  concentration  of  Mg++  ion  in  the  solution?  To 
determine  this  filter  the  solution  and  to  the  clear  filtrate  add  5  cc.  of 
M/10  sodium  phosphate.  In  the  presence  of  magnesium  ion  a  crys- 
talline precipitate  of  MgNH4P04  will  result.  Was  much  magnesium 
ion  still  present?  Making  use  of  the  following  equilibria, 

MgS04<=±Mg+++SO4 
NH4OH  <=>  OH-+NH4+ 

IT 

Mg(OH)2  (non-ionized) 

IT 

Mg(OH)2     (solid) 

note  what  ions  accumulate  as  Mg(OH)2  is  precipitated  by  NH4OH,  and 
what  effect  this  accumulation  has  on  the  concentrations  (Mg++)  and 
(OH~).  Which  one  will  be  affected  the  more,  relatively?  Explain  now 
why  the  precipitation  stopped  when  much  magnesium  was  still  present. 

Procedure  B. — Suppose  we  were  to  mix  20  cc.  each  of  molar  NH4OH 
and  M/5  MgS04,  having  previously  added  to  one  solution  enough  solid 
ammonium  chloride  to  make  the  solution,  after  mixing,  molar  with  respect 
to  this  salt.  Would  a  precipitate  of  Mg(OH)2  be  formed?  (The  solu- 
tions after  mixing  total  40  cc.,  which  equals  0.04  liter.  We  shall,  there- 
fore, need  0.04  mole  of  NH4C1.) 

Molar  NH4C1  is  74  per  cent  ionized.  Calculate  the  total  NH4+ 
concentration;  and  then,  applying  the  principle  of  the  common  ion 
effect,  calculate  the  concentration  of  OH~  which  must  stand  in  equi- 
librium with  this.  From  this  and  the  known  concentration  of  Mg++ 
calculate  the  ion  product  (Mg++)  X  (OH~)2.  You  are  then  able  to 
decide  whether  precipitation  of  Mg(OH)2  will  occur.  Having  made  the 
calculations  perform  the  actual  experiment. 

Exp.  49.  The  Silver-Ammonium  Complex. 

Procedure  A. — Place  10  cc.  of  N/5  silver  nitrate  in  a  small  beaker 
and  titrate  rapidly  with  N/5  ammonium  hydroxide  until  the  precipitate 
first  formed  is  redissolved,  leaving  nothing  more  than  the  faintest 
opalescence.  Note  the  number  of  cubic  centimeters  of  ammonia  solu- 
tion required  to  do  this. 

Judging  from  the  titration  values,  should  the  complex  take  the  form 
AgNHs,  Ag(NH3)2,  or  what?  Is  the  complex  an  anion,  or  a  cation? 
Write  the  formula  of  the  salt  of  the  complex  formed  in  the  above  reac- 
tion. Also  write  an  equation  indicating  the  complete  reaction  between 
the  silver  nitrate  and  the  ammonium  hydroxide.  What  is  the  precipitate 
formed  at  first? 


78  COMPLEX  EQUILIBRIUM 

Procedure  B. — Calculate  whether  a  precipitate  of  silver  chloride 
should  be  formed  if  we  were  to  add  1  cc.  N/10  sodium  chloride  solution 
to  10  cc.  of  the  silver-ammonium  nitrate  solution,  proceeding  as 
follows  : 

The  dissociation  constant  for  the  complex  has  the  value  indicated 
thus: 

(Ag+)X(NH3)* 

(Ag(NH3)2+) 

The  original  concentration  of  the  silver  salt  was  0.2.  If  the  final  volume 
of  the  solution  after  titration  is  three  times  as  great  as  the  original  volume 
and  each  complex  molecule  contains  one  silver  atom  the  concentration 
of  the  complex  salt  is  1/3  of  0.2,  or  0.0666.  Like  any  other  salt,  this 
complex  salt  is  at  this  dilution  about  90  per  cent  ionized.  Calculate 
from  this  the  concentration  of  the  complex  ion  Ag(NH3)2+. 

The  small  size  of  the  constant  shows  that  the  complex  is  only  slightly 
dissociated.  Therefore  we  may  assume  that  the  concentration  of  the 
undissociated  part  is  practically  the  same  as  the  total  concentration  of 
the  complex.  On  the  assumption  that  this  is  true,  substitute  its  value 
in  the  equation  and  calculate  the  concentration  of  the  silver  ion  Ag+. 
[Let  x  =  (Ag+),  then  (NH3)  =2x,  and  we  have  for  the  numerator  of  the 
fraction  x(2x)2.] 

When  1  cc.  of  N/10  NaCl  is  added  to  10  cc.  of  the  nitrate  solution, 
its  concentration  becomes  N/110,  while  that  of  the  complex  is  scarcely 
changed  at  all.  Calculate  the  Cl~  concentration  in  N/110  NaCl, 
assuming  95  per  cent  ionization. 

You  now  have  the  concentration  of 'the  silver  ion  and  the  chloride 
ion  as  they  will  be  at  the  moment  when  you  mix  1  cc.  0.1  N  NaCl  with 
10  cc.  0.0666  N  silver-ammonium  nitrate.  The  solubility  product  for 
AgCl  is  2  X  10~10  at  25°  C.  Find  whether  with  the  above  concentrations 
of  Ag+  and  Cl~,  a  precipitate  of  AgCl  should  be  produced. 

Having  made  the  calculation,  verify  by  actual  experiment. 

Procedure  C. — Will  the  complex  give  a  precipitate  of  AgCl  if  we 
first  add  an  equal  volume  of  N  NH4OH  and  then  add  N/10  NaCl  in 
the  same  proportions  as  in  B? 

Note  that  the  concentration  of  both  the  complex  and  the  NHUOH 
are  reduced  to  half  their  original  concentrations  by  mutual  dilution. 
In  the  case  of  the  latter  we  are  concerned  in  knowing  the  concentration 
of  the  NH3.  The  total  concentration  (NH3+NH4OH)  becomes  0.5 
after  mixing,  and  the  concentration  of  the  NH3  may  be  taken  as  twice 
that  of  NH4OH.  From  this  (NH3)  =2/3  of  0.5,  or  0.33. 

Letting  x  again  stand  for  (Ag+),  we  have  for  the  numerator  of  the 


THE  FERRIC  OXALATE  COMPLEX  79 

fraction  z(0.33)2.  Making  the  proper  substitution  for  the  denominator 
the  value  of  x  may  be  calculated  at  once. 

Having  thus  determined  the  concentration  of  the  Ag4"  ion,  again 
apply  the  solubility  product  principle  and  determine  whether  a  precip- 
itate should  be  obtained. 

Verify  by  actual  experiment,  starting  with  10  cc.  of  the  silver-ammo- 
nium nitrate  solution,  adding  first  an  equal  volume  of  N  NH4OH  and 
then  2  cc.  of  N/10  NaCl. 

Calculate  also  whether  6  cc.  of  N/10  NaCl  should  give  a  precipitate 
of  AgCl,  counting  the  Cl~  ion  concentration  three  times  that  produced 
by  the  addition  of  2  cc.  Prove  this  also  by  experiment;  that  is,  add 
4  cc.  more  to  the  solution  just  tested. 

Procedure  D.  —  Would  a  precipitate  of  silver  bromide  be  produced  if 
KBr  were  substituted  for  NaCl  in  (C)? 

The  concentration  of  the  Br~  ion  may  be  taken  as  equal  to  that  of 
the  Cl~  ion  as  calculated  in  (C).  The  solubility  product  for  AgBr  is 
4.4X10-13. 

When  you  have  decided  whether  a  precipitate  should  be  produced, 
verify  by  adding  1  cc.  N/10  KBr  solution  to  10  cc.  of  the  mixture  of  1 
vol.  silver-ammonium  nitrate  and  1  vol.  normal  NEUOH. 

Exp.  60.  —  The  Ferric-oxalate  Complex. 

When  ferric  sulphate  and  ammonium  oxalate  are  brought  together 
the  complex  ammonium  ferric-oxalate  is  formed  according  to  the  equa- 
tion. 

Fe2(S04)3+6(NH4)2C204.2H20 


+3(NH4)2SO4  +  12H2O 

It  will  be  noted  that  the  weights  of  the  reacting  substances  are  nearly  in 
the  proportion  of  1  :  2.  To  prepare  the  complex,  therefore,  weigh  out 
2  gm.  ferric  sulphate  and  4  gm.  ammonium  oxalate,  grind  together  in  a 
mortar,  and  dissolve  in  100  cc.  of  water  without  heating.  Make  the 
following  tests  with  this  solution: 

(A).  —  To  10  cc.  add  5  cc.  10  per  cent  solution  of  ammonium  thio- 
cyanate.  Do  you  obtain  a  test  for  ferric  ion? 

Test  another  portion  with  ammonium  hydroxide.  What  is  the  pre- 
cipitate produced?  Was  ferric  ion  present?  Point  out  the  reason  why 
these  two  tests  give  different  results. 

To  another  10  cc.  add  3  cc.  6  N  HC1  and  then  5  cc.  thiocyanate  solu- 
tion. Explain  the  result. 

(B).  —  Test  a  small  portion  of  the  solution  for  oxalate  ion  by  adding  a 
solution  of  calcium  chloride.  Is  oxalate  ion  present? 


80  COMPLEX  EQUILIBRIUM 

In  another  10-cc.  portion  dissolve  a  piece  of  solid  ferric  chloride  the 
size  of  a  pea,  and  then  test  for  oxalate  ion  as  before.  Explain. 

Exp.  51.  Amphoteric  Nature  of  the  Aluminum  Group. 

Procedure  A.  Aluminum. — To  5  cc.  aluminum  sulphate  solution  add  a 
dilute  solution  of  NaOH,  drop  by  drop,  with  stirring,  until  a  piece  of 
red  litmus  paper  placed  in  the  solution  barely  shows  a  permanent  blue 
color.  The  precipitate  is  aluminum  hydroxide  Al(OH)s. 

To  show  the  amphoteric  nature  of  this  hydroxide,  treat  one  portion 
of  the  above  mixture  with  an  excess  of  dilute  NaOH  and  another  with  an 
excess  of  dilute  HC1.  Write  equilibria  to  show  the  amphoteric  ioniza- 
tion  of  Al(OH)s,  and  then  show  how  the  equilibria  are  disturbed  by 
adding  NaOH  and  HC1  respectively.  Name  the  two  ions  containing 
aluminum. 

To  the  above  acid  solution  add  an  excess  of  dilute  NH4OH.  What  is 
the  precipitate?  Show  how  the  equilibria  are  disturbed,  and  extend  to 
include  the  solid  phase  (the  precipitate).  To  the  corresponding  alkaline 
solution  add  a  solution  of  ammonium  chloride.  Show  how  the  equi- 
libria are  disturbed  in  this  case. 

Procedure  B.  Chromium. — To  5  cc.  chromium  sulphate  solution  add 
NaOH  as  in  Procedure  A,  taking  care  not  to  use  an  excess.  Treat 
separate  portions  of  the  mixture  thus  produced  with  NaOH  and  HC1  as  in 
A.  Write  equilibria  here  also,  and  show  how  they  are  disturbed  by  acid 
and  alkali.  Name  the  ions  of  chromium  here  involved. 

Precipitate  Cr(OH)3  as  above,  place  in  a  small  casserole,  add  about 
1  gm.  of  sodium  peroxide,  Na2O2,  and  heat  to  boiling  until  the  green 
color  gives  place  to  a  clear  yellow.  What  does  the  yellow  solution  con- 
tain? What  sort  of  reaction  was  this  by  which  chromium  ion,  Cr+++, 
was  changed  to  chromate  ion,  CrO4=?  Would  aluminum  ion  act  in 
the  same  way?  Does  chromium  in  this  hexavalent  form  ever  exhibit 
any  basic  properties;  in  other  words,  is  hexavalent  chromium  ampho- 
teric? Write  the  formula  of  a  hypothetical  hydroxide  of  hexavalent 
chromium.  How  is  this  related  to  chromic  acid,  H2CrO4?  (Note  also 
the  change  of  Al(OH)s  to  HA1O2.)  How  can  hexavalent  chromium  be 
changed  back  to  the  trivalent  form? 

Procedure  C.  Iron. — Precipitate  ferric  hydroxide  from  a  solution  of 
the  nitrate  or  chloride,  proceeding  as  in  A.  Treat  separate  portions  of 
the  mixture  with  NaOH  and  HC1  respectively.  Is  ferric  hydroxide 
amphoteric?  Write  equilibria  showing  its  method  of  ionization  and 
the  effect  of  H+  ion,  not  forgetting  the  heterogeneous  equilibrium 
involved. 

Ferric  hydroxide  cannot  be  oxidized  by  the  method  used  for  chro- 


AMPHOTERIC  NATURE  OF  THE  ALUMINUM  GROUP  81 

mium.  To  show,  however,  that  iron  does  possess  latent  acidic  prop- 
erties, proceed  as  follows:  Take  1  gm.  of  ferric  oxide,  mix  with  4  gm.  of 
sodium  peroxide  in  a  small  nickel  crucible,  and  heat,  gently  at  first, 
finally  to  faint  redness.  Cool  the  melt  completely  and  dissolve  out  of 
the  crucible  by  placing  in  50  cc.  of  ice-water  contained  in  a  small  casse- 
role. 

The  purple  color  of  the  solution  is  due  to  the  presence  of  sodium 
ferrate,  Na2FeO4.  The  iron  is  evidently  hexavalent,  and  the  salt  is 
related  to  the  hypothetical  hydroxide,  Fe(OH)6,  just  as  Na2CrO4  is 
related  to  hypothetical  Cr(OH)e.  (Point  out  this  relationship.)  Write 
equilibria  showing  the  acidic  and  possible  basic  ionization  of  Fe(OH)e, 
or  better  the  anhydride  form,  FeO2(OH)2.  Which  mode  of  ionization  is 
more  pronounced? 

The  decomposition  of  the  excess  of  Na202  makes  the  solution  very 
basic.  Could  basic  ionization  of  the  above  hydroxide  occur  in  such 
a  solution?  Could  basic  ionization  occur  in  an  acid  solution?  Sodium 
ferrate  is  very  unstable  at  best,  but  it  has  its  greatest  stability  in  a 
strongly  basic  solution.  Dilute  a  portion  of  the  concentrated  solution 
with  10  volumes  of  water  and  let  stand  for  some  minutes.  Acidify 
another  small  portion,  and  note  the  extreme  rapidity  with  which  the 
basic  ion  is  reduced.  The  difference  in  the  rapidity  of  reduction  in  acid 
and  alkaline  solution  is  very  good  evidence  that  the  two  modes  of 
ionization  actually  exist. 

The  most  stable  ferrate  is  probably  that  of  barium,  and  the  stability 
in  this  case  is  probably  due  mainly  to  insolubility  and  consequent  lack 
of  ionization.  It  would  be  interesting  to  prepare  this  salt.  To  do  so, 
carefully  pour  off  the  purple  solution  from  the  ferric  oxide  beneath  it, 
and  to  this  solution  add  barium  chloride  solution  as  long  as  a  precipitate 
is  produced.  Let  the  precipitate  of  BaFeCU  settle,  and  decant  the  liquid. 
Add  water  and  filter  on  a  Biichner  funnel,  washing  the  precipitate 
thoroughly.  Dry,  and  then  scrape  from  the  paper  and  preserve  as  a 
preparation. 

Exp.  52.  Amphoteric  Nature  of  the  Halogens. 

Procedure  A.  Displacement  of  Negative  Iodine  by  the  More  Nega- 
tive Bromine. — To  5  cc.  of  potassium  iodide  solution  add  5  cc.  of  bromine 
water.  Iodine  is  displaced.  Write  the  ionic  equation  showing  this. 

Procedure  B.  Displacement  of  Positive  Bromine  by  the  More 
Positive  Iodine. — Weigh  out  on  the  laboratory  balance  0.8  gm.  of 
potassium  bromate,  KBrOs,  and  0.6  gm.  of  solid  iodine.  Grind  the  two 
solids  together  in  a  mortar,  transfer  to  a  small  flask,  and  add  10  cc.  of 
water.  Heat  gently  (not  to  boiling)  in  a  hood  with  good  draft.  Is 


82  COMPLEX  EQUILIBRIUM 

bromine  displaced?    Write  an  equation  indicating  the  change.    The 
product  is  potassium  iodate. 

Bromine  in  KBrOs  or  HBrOs  is  pentavalent  positive.  Write  the 
formula  of  a  hydroxide  of  this  bromine.  How  is  HBrOs  related  to  this? 
Write  equilibria  showing  both  basic  and  acidic  ionization  of  the  penta- 
hydroxide  or  its  anhydride  form.  If  the  basic  ionization  were  encour- 
aged, the  pentavalent  ion  Br+++++  would  be  produced.  What  effect 
would  this  have  on  the  speed  of  displacement  by  iodine?  Weigh  out 
a  sample  of  KBrOs  and  iodine  as  above,  grind  together,  and  divide  into 
two  nearly  equal  portions.  Place  these  in  separate  flasks,  and  add  to 
each  5  cc.  of  water.  To  one  then  add  1  cc.  of  6  N  sulphuric  acid.  Let 
the  two  stand  side  by  side  for  a  time  without  heating,  having  them 
properly  labeled  to  prevent  confusion.  Which  one  shows  the  greater 
speed  of  reaction?  Explain  fully. 


CHAPTER  XV 


ELECTROCHEMISTRY 

Exp.  53.  Determination  of  the  Faraday. 

Apparatus. — Copper  coulometer,  milliammeter,  150-ohm  rheostat. 
The  plates  of  the  coulometer  are  of  copper, 

24  gauge,  each  having  a  surface   of   about 

25  sq.    cm.,    counting   both   sides.     Before 
being  used   the    first  time  they  should  be  L 
rubbed  bright  with  emery  paper  and  washed 
with  alcohol  to  remove  grease.     After  this 
the  surfaces  which  go  below  the  solution 
should  not  be  touched.     The  arrangement 
of  the  cell  is  seen  in  Fig.  24.     The  arms  of 
the  plates  have  forked  ends  which  enable 
them  to  be  easily  attached  or  removed. 

Procedure. — The    solution  used  in    the 
coulometer  is  made  up  according  to  the  following  formula : 

PureCuSO4-5H20 75gm. 

H2SO4... 25    " 

Alcohol.  : 25    " 

Water 500    ' 

It  is  best  to  first  dissolve  the  copper  sulphate  by  grinding  with  a 
little  water  and  gradually  adding  more  until  it  is  all  in  solution,  and 
then  to  add  the  acid  and  alcohol.  The  solution  should  be  filtered  unless 
perfectly  clear.  After  each  experiment  the  solution  should  be  returned 
to  the  stock  bottle.  It  may  be  used  over  and  over  indefinitely. 

The  determination  is  conducted  as  follows:  Fill  the  cell  about  half  full 
of  the  copper  solution,  or  so  full  that  the  wide  part  of  the  electrodes  is 
covered,  and  connect  in  series  with  a  rheostat  (150  ohms  in), an  ammeter, 
and  two  storage  cells  (4  volts) .  Be  sure  that  you  know  which  way  the 
current  is  running,  and  leave  one  connection  open  until  the  apparatus 
is  inspected  by  the  instructor  (note  1).  When  everything  is  ready  close 
the  circuit,  and  then  adjust  the  rheostat  so  as  to  give  a  current  of  about 
0.25  ampere  (note  2).  Now,  leaving  all  the  connections  and  adjust- 

83 


84 


ELECTROCHEMISTRY 


ments  undisturbed,  remove  the  cathode  and  wash  it,  first  with  water 
and  then  with  alcohol.  Dry  by  pressing  between  two  filter  papers  and 
then  warming  gently  by  holding  high  over  a  small  flame.  Finally  cool 
and  weigh  accurately. 

Now  return  the  cathode  to  its  place,  and  just  at  the  moment  when  it  is 
plunged  into  the  solution  take  the  time  to  a  fraction  of  a  minute.  (It 
will  be  best  to  start  exactly  on  the  minute.)  The  ammeter  reading 
should  be  the  same  as  before,  but  it  should  be  retaken  and  adjusted  if 
necessary,  and  the  value  then  recorded. 

After  about  twenty-five  minutes  (note  the  exact  time)  break  the  con- 
nection, and  then  quickly  wash  and  dry  the  cathode  as  before  (note  3). 
Finally  weigh  the  cathode  and  determine  the  weight  of  the  copper 
deposited. 

Calculate  the  number  of  seconds  during  which  the  current  was 
running;  and  then,  remembering  that  amperes  multiplied  by  seconds 
give  coulombs,  calculate  the  number  of  coulombs  of  electricity  used. 
This  was  the  quantity  of  electricity  required  to  deposit  the  copper  you 
obtained.  Calculate  first  the  copper  equivalent  of  the  coulomb,  and 
then  the  number  of  coulombs  required  to  deposit  one  equivalent  weight 
of  copper.  The  latter  is  the  faraday.  The  standard  value  for  the  former 
is  0.000329, 

NOTES. — (1)  The  ammeter  may  be  ruined  by  careless  work.  By  having  a  high 
resistance  in  the  instrument  when  first  turning  on  the  cur- 
rent, you  will  prevent  trouble. 

(2)  If  too  heavy  a  current  is  used  the  copper  ions  will  not 
be  able  to   carry   it  alone,    and   some   hydrogen  will  be 
deposited;  that  is,  too  little  copper  will  be  obtained.     If 
too  small  a  current  is  used,  some  of  the  copper  ions  will  only 
partially  discharge,  giving  cuprous  ions  Cu+.     This  again 
gives  too  little  copper. 

(3)  Copper  is  slowly  soluble  in  a  solution  of  cupric  ion, 
forming  cuprous  ion;  hence  the  washing  should  be  accom- 
plished as  quickly  as  possible. 

Exp.  54.  Electrode  Reactions. 

Apparatus. — A  6-in.  U-tube  with  two  platinum 

electrodes    and   one    copper    electrode    (Fig.    25). 

This  apparatus  will  be  used  for  all  the  following 

experiments. 

Procedure  A.  Electrolysis  of  Hydrochloric  Acid 

with  Platinum  Electrodes. — Fill  the  U-tube  with 

normal  HC1  until  the  electrodes  are  just  covered. 
Electrolyze,  using  8  volts.  After  about  five  minutes  raise  the 
cathode  and  instantly  apply  a  lighted  match  to  the  mouth  of  the 


MIGRATION  VELOCITY  OF  HYDROGEN  85 

tube.  Explain.  Cautiously  waft  a  little  of  the  anode  gas  to  the  nos- 
trils. What  is  it? 

Show  that  one  of  the  electrode  reactions  is  reduction  and  the  other 
oxidation. 

Do  not  allow  this  electrolysis  to  proceed  long.  The  reason  is 
obvious. 

Procedure  B.  Electrolysis  of  Hydrochloric  Acid  with  Copper  Anode 
and  Platinum  Cathode. — Make  the  proper  change  in  the  electrodes,  and 
then  proceed  as  in  A.  Is  chlorine  gas  evolved?  After  ten  minutes 
remove  a  little  of  the  anode  liquid  and  add  to  it  an  excess  of  ammonium 
hydroxide.  What  ion  was  present?  Show  exactly  how  it  came  there. 
(Do  not  dodge  the  question  by  saying  that  "  copper  was  dissolved  off  the 
anode.") 

Procedure  C.  Electrolysis  of  Sulphuric  Acid  with  Platinum  Elec- 
trodes.— Arrange  the  U-tube  and  electrodes  as  in  A  and  electrolyze 
normal  sulphuric  acid,  using  12  volts.  Test  the  electrode  products 
(gases).  What  are  they?  Explain  their  source.  How  would  the 
results  differ  if  a  copper  anode  were  used? 

Procedure  D.  Electrolysis  of  Sodium  Sulphate  with  Platinum 
Electrodes. — Electrolyze  a  normal  solution  of  sodium  sulphate,  using 
10  volts.  What  gases  are  evolved?  Explain  carefully  their  source. 
Test  the  cathode  liquid  with  red  litmus  paper  and  the  anode  liquid  with 
blue.  Explain. 

Procedure  E.  Pole  Indicator. — Moisten  a  piece  of  red  litmus  paper 
with  a  solution  of  sodium  chloride  and  place  upon  it,  1  cm.  apart,  the 
"  lead  "  wires  from  a  storage  cell.  At  one  pole  the  paper  turns  blue,  at 
the  other  it  is  bleached  white.  Which  pole  is  positive  and  which  is 
negative?  Explain. 

Procedure  F.  Electrolysis  of  Ferric  and  Ferrous  Salts. — Electrolyze 
a  solution  of  ferric  chloride,  using  platinum  electrodes.  After  five 
minutes,  test  the  cathode  liquid  for  Fe++  by  means  of  ferricyanide 
solution.  Explain.  Some  metallic  iron  may  be  deposited.  Explain. 
Test  for  it  by  immersing  the  cathode  in  warm  6  N  HC1.  Evolution  of 
hydrogen  indicates  a  metal,  which  must  be  iron. 

Fill  the  U-tube  with  a  solution  of  ferrous  sulphate,  and  electrolyze  as 
above.  After  five  minutes  test  for  ferric  iron.  (Where?)  Use  ammo- 
nium thiocyanate.  What  happens  in  this  case  at  the  other  electrode? 

Exp.  55.  Migration  Velocity  of  Hydrogen  and  Hydroxyl  Ions. 

Apparatus. — Same  as  in  Exp.  52. 

Procedure. — Prepare  a  2  per  cent  stock  solution  of  agar-agar  as 
follows:  Take  5  gm.  of  the  powdered  material,  grind  to  a  smooth,  thin 


86  ELECTROCHEMISTRY 

paste  with  a  small  amount  of  water,  then  stir  into  it  250  cc.  of  boiling 
water,  and  boil  for  about  two  minutes. 

Take  60  cc.  of  the  agar  solution  (still  hot),  add  to  it  15  cc.  of  a  sat- 
urated solution  of  potassium  nitrate,  10  drops  of  phenolphthalein  indi- 
cator, and  10  drops  of  N/2  NaOH.  Mix  thoroughly  by  pouring  back 
and  forth  between  two  beakers  several  times.  Now  divide  into  two 
equal  parts,  and  to  one  part  add  N/2  NHOs  from'  a  burette  (about 
4  drops),  until  the  color  of  the  indicator  is  just  discharged  after  thorough 
stirring,  and  then  add  exactly  the  same  amount  in  excess. 

You  now  have  two  solutions,  both  of  which  will  become  solid  jellies 
on  cooling,  both  of  which  contain  the  same  electrolyte  (KNO.s),  both 
containing  the  same  indicator,  one  containing  a  small  definite  excess  of 
H+  ion,  and  the  other  the  same  excess  of  OH~  ion. 

Pour  into  the  U-tube  just  enough  of  the  alkaline  solution  to  form  a 
seal  at  the  bottom;  incline  the  tube  so  that  one  end  of  the  solution  comes 
near  the  center,  and  then  cause  the  jelly  to  set  by  running  cold  water 
over  the  tube.  When  the  jelly  is  solid  enough  to  stand  up  well,  fill  the 
shorter  arm  of  the  tube  with  the  alkaline,  and  the  longer  with  the  acid, 
solution,  stopping  in  each  case  about  2  in.  from  the  top  of  the  tube. 
Now  clamp  the  U-tube  upon  the  ring  stand  and  let  the  jelly  set  solid. 
Finally  fill  the  end  next  the  acid  jelly  with  N/2  NaOH  and  that  next  the 
alkaline  with  N/2  HNOs,  insert  the  electrodes,  and  immediately  turn 
on  the  current  (16  volts),  making  the  electrode  in  the  NaOH  the  cathode. 
Take  the  time  and  allow  the  action  to  proceed  for  one  hour. 

You  notice  that  the  OH~  ion  from  the  NaOH  moves  slowly  towards 
the  anode,  coloring  the  indicator  pink,  while  the  H+  ion  from  the  HNOs 
travels  towards  the  cathode,  removing  the  color  of  the  indicator. 

After  one  hour,  measure  the  distances  passed  over  by  each  of  the 
two  ions.  If  the  tube  had  been  left  for  one  hour  without  turning  on  the 
current,  the  acid  would  have  diffused  over  a  distance  of  1.5  cm.  and  the 
alkali  1.1  cm.  (This  has  been  determined  by  experiment.)  In  calcu- 
lating the  distances  passed  over  in  one  hour  through  the  influence  of  the 
applied  potential,  we  must,  therefore,  subtract  these  amounts  from  the 
observed  distances.  Having  done  this,  measure  the  average  dis- 
tance between  the  electrodes  around  the  bend  of  the  U-tube,  and  then 
calculate  the  potential  gradient  used.  Calculate  also  what  would  have 
been  the  distances  covered  by  the  ions  if  the  potential  gradient  had  been 
1.  The  standard  values  at  18°  C.  are  H+  11.52  cm.  per  hour,  and  OH~ 
6.48  cm.  per  hour.  Your  values  will  be  somewhat  lower  than  these, 
because  of  the  retarding  influence  of  the  jelly. 


MIGRATION  OF  A  COMPLEX  ION 


87 


Exp.  56.  Migration  of  a  Complex  Ion. 

Apparatus. — The  specially  constructed  U-tube  which  is  seen  in 
Fig.  26,  and  which  permits  one  liquid  to  be  run  under  another  without 
mixing.  When  in  use  the  apparatus  is  supported  by 
a  ring  stand,  the  clamp  grasping  one  of  the  arms  of 
the  U,  not  the  small  tube  at  the  back.  The  corks 
supporting  the  electrodes  are  grooved  to  allow  the 
escape  of  gases. 

Procedure. — Prepare  25  cc.  of  a  solution  of  cupric 
bromide  containing  1  gm.  CuBr2  to  2.5  gm.  water. 
Having  the  apparatus  properly  supported  and  the 
electrodes  in  place,  fill  the  thistle  tube  with  the  solu- 
tion, taking  care  that  the  air  is  all  out  of  the  connect- 
ing tube  above  the  stop-cock;  then  cautiously  let 
the  solution  flow  around  until  it  just  reaches  the 
bottom  of  the  U.  Now  place  in  the  U-tube  enough 
normal  nitric  acid  to  stand  about  1  inch  high  in  the 
arms,  and  then  cautiously  let  the  copper  solution  in 
under  the  acid  until  the  acid  at  the  top  just  covers  the 
electrodes. 

By  means  of  a  narrow  gummed  label,  mark  the 
boundary  line  of  the  brown  solution  on  the  side  which  is  to  be  the  cathode 
compartment,  and  then  turn  on  the  current  (10  volts).  Allow  the 
action  to  proceed  for  half  an  hour. 

In  what  direction  does  the  copper  ion  move?  Explain.  In  what 
direction  does  the  boundary  line  of  the  brown  move?  If  the  brown 
color  were  due  to  undissociated  CuBr2  would  it  move  at  all?  Why? 
To  what  may  the  brown  color  be  due? 

Exp.  57.  The  Daniell  Cell. 

Apparatus. — A  glass  tumbler  of  table  size.  A  porous  battery  cup 
4X8  cm.  A  3-hole  rubber  stopper  No.  7.  A  piece  of  24-gauge  sheet 
copper  cut  as  seen  in  Fig.  27.  Two  zinc  rods  10  cm.X6  mm.  Volt- 
meter and  ammeter  of  low  range.  Copper  wire  and  connectors. 


FIG.  26. 


16cm. 


FIG.  27. 


88  ELECTROCHEMISTRY 

If  the  copper  is  taken  in  sheets  of  the  size  and  shape  shown  in  A, 
two  electrodes  can  be  made  from  one  piece  without  any  waste  of  material. 
The  copper,  after  being  cut,  is  rolled  into  a  cylinder  to  fit  the  tumbler, 
and  the  arm  is  bent  as  seen  in  B  to  hold  the  electrode  firmly  upright. 
The  zinc  rods  are  inserted  in  the  stopper  as  seen  in  C,  and  are  connected 
to  the  copper  '  lead  "  wire. 

Procedure  A.  E.M.F. — Having  assembled  the  pieces  of  appa- 
ratus seen  above,  fill  the  porous  cup  about  three-fourths  full  of  normal 
zinc  sulphate  solution  and  insert  the  stopper  carrying  the  zinc  electrodes. 
Place  the  copper  electrode  in  the  tumbler,  set  the  porous  cup  inside  it, 
and  then  add,  in  the  outer  compartment,  sufficient  saturated  solution  of 
copper  sulphate  to  cover  the  electrode. 

You  are  first  to  measure  the  E.M.F.  of  the  cell;  but  before  doing  this 
decide  from  which  electrode  the  current  will  come,  and  if  your  volt- 
meter has  more  than  one  range,  decide  which  range  to  use.  Connect 
the  electrode  from  which  the  positive  current  comes  (the  cathode)  to  the 
side  of  the  voltmeter  marked  + ,  and  the  other  to  the  binding  post  giving 
the  proper  range.  Always  be  very  careful  not  to  connect  any  electrical 
measuring  instrument  backwards,  or  to  use  it  with  a  current  which  is 
beyond  its  range. 

When  you  are  sure  about  these  two  points,  connect  the  voltmeter 
and  read  and  record  the  voltage  of  the  cell.  How  does  this  voltage 
compare  with  the  calculated  value  for  the  couple,  Zn/N-Zn++  — 
N  •  Cu++/Cu?  Should  it  be  just  the  same?  Why? 

Procedure  B.  Internal  Resistance. — Connect  the  cell  with  an  amme- 
ter, taking  the  precautions  noted  above,  and  carefully  take  the  reading. 
According  to  Ohm's  law,  C=E/R,  or  R=E/C.  The  copper  wires 
and  the  ammeter  offer  practically  no  resistance,  so  the  resistance  R  of 
the  circuit  is  practically  all  inside  the  cell.  Substitute  for  E  the  value 
found  in  A ,  and  for  C  the  present  reading  of  the  ammeter,  and  calculate 
R.  Upon  what  does  the  internal  resistance  depend? 

Procedure  C.  Effect  of  Concentration. — Remove  the  solutions  from 
the  cell,  saving  only  the  copper  solution,  and  substitute  for  them  the 
following :  For  the  dilute  zinc  sulphate  solution  substitute  a  very  con- 
centrated solution  of  zinc  chloride  (note  1),  and  for  the  concentrated 
copper  sulphate  substitute  one  of  N/100  concentration. 

Now  assemble  the  apparatus,  as  before,  and  take  the  voltage. 
Explain  the  change. 

Add  enough  ammonium  hydroxide  to  the  copper  solution  to  throw 
the  copper  into  the  blue  complex,  Cu(NH3)4++.  Note  the  effect  on 
the  voltage.  Explain. 

Finally  add  somewhat  more  than  enough  potassium  cyanide  solution 


A  CONCENTRATION  CELL  89 

to  remove  the  blue  color  of  the  copper-ammonia  complex.  The  copper 
is  now  still  more  completely  locked  up  than  before  by  being  thrown  into 
the  extremely  stable  complex,  Cu(CN)  2  ~.  (Caution:  Potassium  cyanide 
is  dangerous;  keep  it  off  your  hands,  and  also  off  the  desk.)  Note  the 
effect  on  the  voltage,  reversing  the  connections  if  necessary.  Explain. 

When  through  with  the  experiment,  return  the  zinc  chloride  solution 
to  the  bottle  and  carefully  wash  out  the  apparatus.  Pour  the  cyanide 
solution  into  the  sink  and  immediately  wash  it  down  with  water.  Leave 
the  apparatus  filled  with  water. 

NOTE. — Zinc  chloride  is  used  simply  because  of  its  great  solubility. 

Exp.  58. — A  Concentration  Cell. 

Apparatus. — Same  as  for  the  Daniell  cell,  excepting  that  a  zinc 
electrode  is  substituted  for  the  copper. 

Procedure. — Place  in  the  porous  cup  the  concentrated  zinc  chloride 
solution  used  above,  and  in  the  outside  compartment  use  a  solution  of 
any  zinc  salt  of  N/100  concentration.  Decide  which  way  the  current 
will  flow,  then  connect  in  the  proper  way  and  take  the  voltage.  Explain 
the  source  of  the  E.M.F.  and  the  direction  of  the  current. 

Leave  the  apparatus  filled  with  water. 

Exp.  59.  Decomposition  Voltage. 

A.  Decomposition  Voltage  of  Sulphuric  Acid  with  Platinum  Elec- 
trodes. 

Apparatus. — The  coulometer  cell  used  for  Faraday's  law.  A  pair 
of  platinum  electrodes  of  the  same  size  and  shape  as  the  copper.  A 
voltmeter  and  ammeter,  both  of  low  range.  A  rheostat  of  range  2200 
ohms,  0.5  ampere. 

Procedure. — Set  the  platinum  electrodes  in  place,  taking  great  care 
not  to  bend  them,  and  fill  the  cell  to  the  proper  depth  with  normal 
sulphuric  acid  (note  1).  Connect  with  voltmeter,  ammeter,  rheostat 
(2200  ohms  in),  and  six  storage  cells  (12  volts). 

Now  slowly  turn  out  the  resistance  until  the  voltmeter  reads  0.6, 
and  then  read  the  ammeter.  Increase  the  voltage  to  0.8,  and  again  read 
the  ammeter.  Continue  in  this  way  until  a  voltage  of  about  1.6  is 
reached,  and  then  make  the  intervals  smaller  for  a  few  readings.  Note 
the  voltage  where  a  marked  increase  in  the  amperage  begins.  This  is 
the  decomposition  voltage.  Take  a  few  readings  after  the  change  is 
noted,  to  complete  the  list,  but  do  not  let  the  amperage  go  above  0.5. 

Plot  the  values  obtained  on  coordinate  paper,  making  voltages 
abscissa?  and  amperages  ordinates,  and  then  draw  a  smooth  curve. 
The  decomposition  point  can  then  be  plainly  seen. 


90 


ELECTROCHEMISTRY 


Mercury 


FIG.  28. 


NOTE. — Normal  sulphuric  acid  does  not  give  normal  ion  concentration,  for  it  is 
only  about  0.5  ionized;  but  this  slight  difference  has  no  appreciable  effect  on  the 
decomposition  voltage. 

B.   Decomposition    Voltage  of   Sulphuric  Acid  with    Mercury 

Cathode. 

Apparatus. — Same  as  in  A,  excepting  that  a  copper  connecting  wire, 
which  reaches  from  the  binding  post  on  the  cell  cover  to  the  bottom  of  the 

cell,  is  substituted  for  one  of  the  platinum 
electrodes.  This  makes  the  connection  with 
the  mercury  cathode.  It  is  rubber-insulated 
where  it  passes  through  the  liquid. 

Procedure. — Place  in  the  cell  a  layer  of 
clean  mercury  just  sufficient  to  cover  the 
bottom,  and  upon  this  place  the  normal 
sulphuric  acid  as  above.  When  the  cover  is 
in  place,  the  foot  should  be  entirely  covered 
with  mercury — no  uninsulated  copper  should 
come  into  contact  with  the  acid.  Proceed 
to  determine  the  decomposition  potential 
exactly  as  in  A,  plotting  the  values  on  co- 
ordinate paper. 

Discuss  the  subject  of  overvoltage,  and  show  its  relation  to  the 
results  of  this  experiment  (for  table  of  hydrogen  potentials  on  various 
metals,  see  below) : 

POTENTIAL  OF  HYDROGEN  ON   VARIOUS   METALS 

(Sign  of  Solution) 

H2  on  Hg +0.51 

H2  on  Zn +0 . 43 

H2  on  Pb +0 . 37 

H2onSn +0.26 

H2  on  Cu -0 . 04 

H2  on  Ni — 0 . 06 

H2  on  Ag -0 . 12 

H2  on  Pt  (smooth) -0 . 18 

H2  on  Fe -0 . 19 

H2  on  Pt  (black) -0.27 

Exp.  60.  Displacement  Reactions. 

Procedure  A.  Action  of  Iron  and  Zinc  on  the  Ions  of  Tin. — Take 
20  cc.  of  stannic  chloride  solution,  add  to  it  2  cc.  of  concentrated  hydro- 
chloric acid,  and  divide  between  two  test-tubes.  In  on  3  put  a  piece  of 
sheet  zinc,  and  in  the  other  a  piece  of  iron  wire  which  has  been  pre- 
viously cleaned  with  emery  paper  and  rolled  into  a  neat  helix.  Warm 


DISPLACEMENT  REACTIONS  91 

both  tubes  to  start  the  reaction,  and  then  set  aside  for  fifteen  minutes. 
The  action  will  be  somewhat  slow,  because  a  solution  of  stannic  chloride 
does  not  contain  much  stannic  ion.  During  the  act  of  solution  most  of 
the  tin  will  have  gone  into  the  form  of  the  complex  anion  SnOs=.  (Ex- 
plain the  effect  of  HC1  on  t  is.) 

After  fifteen  minutes  note  what  has  occurred.  Has  metallic  tin 
(gray,  leathery  flakes)  appeared  in  both  cases?  Explain  its  presence  or 
absence.  What  has  occurred  in  the  solution  containing  the  iron  wire? 
Test  for  stannous  ion  by  the  common  qualitative  reaction.  Explain. ' 

Take  5  cc.  of  stannous  chloride  solution  and  add  a  piece  of  sheet  zinc. 
Can  you  account  or  the  exceedingly  rapid  displacement  of  the  tin? 

Procedure  B.  Selective  Displacement. — Take  3  cc.  of  stannous 
chloride  solution,  3  cc.  of  antimony  trichloride  solution,  and  2  cc.  of 
6  N  HC1.  In  this  m'xture  place  a  coil  of  clean  iron  wire,  heat  if  neces- 
sary to  start  the  reaction,  and  then  leave  for  fifteen  minutes.  At  the 
end  of  this  time  remove  the  wire,  and  carefully  wash  the  precipitate  by 
decantation  until  the  wash  water  no  longer  gives  a  test  for  Sn++.  What 
is  the  precipitate?  From  your  observat'on  in  (A)  would  you  expect  it 
to  contain  tin?  Treat  the  precipitate  with  1  cc.  of  concentrated  HC1. 
Is  there  any  evidence  that  a  metal  is  "  dissolving,"  i.e.,  is  hydrogen  dis- 
placed? After  five  minutes,  dilute  the  acid  solution  with  two  volumes  of 
water,  filter,  and  test  for  tin.  Remove  the  black  residue  to«a  casserole 
and  treat  with  2  cc.  6  N  HC1  and  2  drops  only  of  concentrated  HNOs, 
warming  until  solution  is  complete.  Dilute  with  10  cc.  of  water,  and 
add  a  solution  of  hydrogen  sulphide.  An  orange-colored  precipitate  of 
Sb2Ss  indicates  antimony. 

Why  is  antimony  displaced  by  iron  while  tin  is  not?  Why  did  anti- 
mony not  dissolve  when  the  precipitate  was  treated  with  concentrated 
HC1  as  above?  When  antimony  dissolves  in  aqua  regia,  as  above,  does  it 
displace  anything?  What  does  it  do? 

Procedure  C.  Hydrogen  Displacement  and  Overvoltage. — In  decid- 
ing whether  a  metal  will  "  dissolve  "  in  an  acid,  that  is,  displace  hydrogen 
ion,  it  is  not  sufficient  to  note  that  hydrogen  stands  below  this  metal 
in  the  potential  series.  In  giving  hydrogen  its  position  in  the  potential 
series  we  were  assuming  that  its  potential  was  measured  on  the  surface 
of  a  platinized  platinum  electrode.  On  the  surface  of  any  other  metal, 
its  potential,  and  consequently  its  position  in  the  series,  might  be  widely 
different.  Consult  the  table  above  and  note  where  hydrogen  would  be 
placed  if  its  potential  were  taken  on  the  surface  of  mercury;  on  silver; 
on  lead. 

Place  50  cc.  of  3  N  hydrochloric  acid  in  a  small  beaker.  Into  this 
dip  a  strip  of  clean  sheet  zinc.  Note  that  it  immediately  begins  to 


92  ELECTROCHEMISTRY 

dissolve,  displacing  hydrogen  ion.  Explain.  Is  the  potential  of  hydro- 
gen on  zinc  above,  or  below,  that  of  zinc  itself? 

Remove  the  zinc  from  the  acid  and  bring  it  into  contact  with  a  small 
drop  of  mercury.  Rub  this  over  the  surface  until  the  coating  of  amal- 
gam is  perfectly  homogeneous.  Now  place  the  strip  again  in  the  acid. 
Is  hydrogen  displaced?  Note  that  the  amalgam  on  the  surface  is 
really  a  solution  of  zinc  in  mercury,  so  that  the  zinc  is  still  in  contact 
with  the  acid  as  it  was  before.  But  the  hydrogen  must  now  come  off 
on  a  surface  of  mercury.  Is  the  potential  of  hydrogen  on  mercury 
above,  or  below,  that  of  the  zinc  itself?  (See  table  above.)  Does  this 
account  for  what  you  observe? 

Now,  if  we  offer  another  metal  surface  in  contact  with  the  amal- 
gamated zinc,  where  the  potential  of  the  hydrogen  will  be  lower  than 
that  of  zinc,  we  should  expect  the  displacement  of  hydrogen  by  zinc  to 
proceed.  Referring  to  the  table,  would  you  consider  iron,  or  lead,  to  be 
the  better  for  this  purpose?  Test  by  taking  clean  strips  of  these  metals 
and  holding  them  under  the  acid  in  contact  with  the  amalgamated  zinc. 


APPENDIX  I 

General  Outline  of  Laboratory  Work 

The  following  outline  shows  in  some  detail  the  author's  time  schedule 
covering  the  laboratory  work  for  the  whole  second  year  course: 


First  Semester 


Weeks. 


1 

Registration,  preliminary  laboratory  work,  including  checking  over 
apparatus,  etc. 

2,3,4 

Experiments  1-1  1  inclusive.      Groups  as  outlined  under  Appendix  II 
below. 

5,  6,  7,  8 

Experiments  12-19  inclusive,  all  individual  work. 

9,  10,  11,  12, 
13,  14,  15 

Gravimetric  analysis.    Constant  application  of  principles,  and  prac- 
tice in  calculation. 

16,  17 

Experiments  20-25  inclusive.     Groups  on  24  and  25;  26  optional. 

18 

Examination  week. 

Second  Semester 

1,2,3 

Experiments  27-35  inclusive.     Groups  on  28,  31,  32,  33. 

4,  5,  6,  7,  8,  9, 

Volumetric  analysis.  Constant  application  of  principles,  and  practice 

10 

in  calculation. 

11,  12,  13,  14 

Experiments  32-52  inclusive.  Groups  on  32,  33,  and  41B;  43  optional. 

15,  16 

Experiments  53-59  inclusive.    Groups  as  outlined  under  Appendix 
II  below.     Exp.  60.     Checking  out. 

17,  18 

Examinations  and  commencement. 

It  will  be  noted  that  more  time  seems  to  be  assigned  to  the  theoretical 
work  than  to  the  practical  analytical  work.  This  is  really  not  the  case, 
for  several  of  the  theoretical  experiments  involve  accurate  analytical 
work.  The  time  is  about  evenly  divided  between  the  two  types  of  work. 

93 


APPENDIX  II 
Grouping  of  Students  for  the  Use  of  Special  Apparatus 

The  laboratory  work  covering  certain  topics  requires  special  appa- 
ratus, of  which  only  a  limited  number  of  pieces  will  be  available.  In 
such  cases  the  only  feasible  way  to  proceed  is  to  assign  the  experiments 
in  such  a  way  that  the  smallest  possible  number  of  students  will  on  any 
given  day  be  working  on  the  same  experiment.  There  are  two  groups 
of  experiments  in  the  manual  where  this  procedure  is  particularly 
necessary,  namely  Exps.  1-1 1  and  Exps.  53-59.  The  procedure  for  the 
first  group  has  been  as  follows :  The  first  three  experiments  are  short,  and 
so  may  be  finished  in  one  period  of  three  and  a  half  hours.  The  other 
eight  will  occupy  a  period  each,  if  properly  done.  This  makes  it  possible 
to  do  the  whole  eleven  in  nine  periods,  or  three  weeks,  where  a  student 
puts  in  three  periods  per  week.  To  avoid  confusion  and  induce  students 
to  study  the  experiments  ahead,  it  is  suggested  that  the  instructor 
prepare  a  chart  containing  the  students'  names  to  whom  the  experi- 
ments are  severally  assigned  so  as  to  have  the  smallest  possible  number 
on  any  one  experiment  at  a  time,  and  so  that  each  student  will  be 
doing  the  experiments  in  as  nearly  the  normal  order  as  possible.  The 
following  chart  will  make  this  clearer. 

In  the  course  for  which  the  chart  was  prepared,  the  work  began 
on  Wednesday.  The  stars  indicate  the  days  elected  by  the  stu- 
dent. Thus,  student  A  elected  Monday,  Wednesday  and  Friday, 
student  B,  Monday,  Thursday  and  Friday,  etc.  The  experiments  were 
assigned  to  A,  B,  C,  D  in  the  normal  order.  This  brings  A  and  C  on 
1,  2,  3,  together  and  A,  B,  and  C  on  4  together.  E,  F,  and  G  start  off 
with  4,  H  and  I  with  5,  etc.  If  only  nine  students  are  due  on  Wednes- 
day, each  can  be  given  a  separate  experiment.  If  the  number  is  eighteen 
two  students  will  start  working  on  each  experiment,  etc.  Note  that 
after  a  student  once  begins,  the  experiments  follow  on  in  normal  order 
for  him,  although  he  may  begin  with  number  5  or  6. 

Experiments  53-59  can  be  arranged  to  cover  four  periods,  thus: 
(53-54),  (55-56),  (57-58),  (59).  The  charting  is  done  exactly  as  indi- 
cated above,  and  so  needs  no  comment. 

94 


GROUPING  OF  STUDENTS  FOR  USE  OF  APPARATUS 


95 


Stu- 
dent. 

Wed. 

Thur. 

Fri. 

Mon. 

Tue. 

Wed. 

Thur. 

Fri. 

Mon. 

Tue. 

Wed. 

Thur. 

Fri. 

Mon. 

Tue. 

* 

* 

* 

* 

* 

* 

* 

* 

* 

A 

123 

4 

5 

6 

7 

8 

9 

10 

11 

* 

* 

* 

* 

* 

* 

* 

* 

* 

B 

123 

4 

5 

6 

7 

8 

9 

10 

11 

* 

* 

* 

* 

* 

* 

* 

* 

* 

C 

123 

4 

5 

6 

7 

8 

9 

10 

11 

* 

* 

* 

* 

* 

* 

* 

* 

* 

D 

123 

4 

5 

6 

7 

8 

9 

10 

11 

* 

* 

* 

* 

* 

* 

* 

* 

* 

E 

4 

5 

6 

7 

8 

9 

10 

11 

123 

* 

* 

* 

* 

* 

* 

* 

* 

* 

F 

4 

5 

6 

7 

8 

9 

10 

11 

123 

* 

* 

* 

* 

* 

* 

* 

* 

* 

G 

4 

5 

6 

7 

8 

9 

10 

11 

123 

* 

* 

* 

* 

* 

# 

* 

* 

* 

H 

5 

6 

7 

8 

9 

10 

11 

123 

* 

4 

* 

* 

* 

* 

* 

* 

* 

* 

I 

5 

6 

7 

8 

9 

10 

11 

123 

4 

* 

* 

# 

* 

* 

* 

* 

* 

* 

J 

6 

7 
* 

8 

9 

10 

11 

123 

4 

5 



* 

* 

* 

* 

* 

* 

* 

* 

K 

V.+  n 

6 

7 

8 

9 

10 

11 

123 

4 

5 



1 

Where  single  experiments  are  encountered  which  call  for  special 
apparatus,  the  grouping  can  be  arranged  after  the  students  come  to  the 
laboratory.  Examples  of  this  sort  are  experiments  24,  25,  28,  31,  32,  33, 
4 IB.  All  that  is  necessary  in  these  cases  is  to  see  that  there  are  not  too 
many  working  together.  The  group  will  sometimes  tend  to  rush  through 
the  experiment  and  finish  in  the  middle  of  the  period.  In  such  cases 
they  should  repeat,  and  in  any  case  they  should  not  be  allowed  to  go 
faster  than  the  schedule  requires. 

Attention  may  be  called  here  to  the  general  schedule  outlined  in  the 
preface  to  the  text,  where  the  time  required  for  large  groups  of  experi- 
ments is  indicated.  No  trouble  will  be  experienced  in  actually  getting 
the  experiments  done  in  the  time  specified,  provided  the  time  spent  in 
laboratory  work  is  as  there  indicated  (nine  or  ten  hours  per  week  for 
thirty-two  weeks). 


APPENDIX  III 
Data  and  Suggestions  Regarding  Individual  Experiments 

As  used  by  the  author,  the  course  of  experiments  given  in  this  manual 
is  made  to  occupy  about  half  the  time  of  a  student  working  nine  hours 
per  week  for  thirty-two  weeks.  With  that  arrangement  it  is  necessary 
to  furnish  much  of  the  apparatus  in  a  set-up  form,  and  in  some  cases 
also  to  furnish  standard  solutions.  The  following  notes  indicate  what 
is  furnished  in  this  way,  and  incidentally  give  some  suggestions  about  the 
experiments  themselves.  A  list  of  the  materials  and  reagents  which 
must  be  looked  after  is  also  given  in  each  case  Note  that  where  a 
reagent  has  already  been  prepared,  the  fact  is  indicated  by  a  cross 
reference,  e.g.,  "  as  in  29."  Sample  values  are  given  where  possible,  to 
indicate  what  may  be  expected  in  the  way  of  accuracy, 

Exp.  1.  Brownian  Movement. 

Apparatus. — Microscope  and  slides. 
Chemicals. — Gamboge. 

Exp.  2. — Evaporation  in  Vacuum. 

Apparatus. — Two  vacuum  desiccators  containing  5-cm.  watch  glasses. 
The  air  pump  used  is  a  Cenco-Nelson,  run  by  a  |  H.P.  motor.  The 
pump  and  motor  are  fastened  on  a  wooden  base,  arrangement  being 
made  for  taking  up  the  slack  of  the  belt.  On  the  same  base  are  also  a 
closed  tube  manometer  and  a  trap  to  prevent  oil  from  being  drawn  back 
into  the  apparatus. 

Pressure  tubing,  tied  in  place  with  waxed  cord,  will  be  needed  in 
making  the  connections. 

Results. — Time  1  hour  30  minutes.  In  second  desiccator  0.8  gm. 
left. 

Exp.  3.  Boyle's  Law. 

Apparatus. — The  Boyle's-law  apparatus.  The  mercury  becomes 
dirty  in  time,  probably  on  account  of  sulphur  in  the  rubber  stopper  near 
it.  This  will  adhere  to  the  end  of  the  tube  containing  the  compressed 
air  and  allow  an  occasional  bubble  to  creep  in.  This,  of  course,  will 
slightly  change  the  value  of  PV. 

96 


DATA  AND  SUGGESTIONS  97 

Students  are  very  likely  to  read  the  barometer  in  millimeters  and 
values  on  the  apparatus  in  centimeters  and  then  add  the  figures  directly. 

If  students  are  careless  about  handling  the  tubes,  the  air  may  be 
warmed.  PV  will  then  grow  larger  as  they  proceed  with  the  experiment. 

Results.— PV  =  1712,  1713,  1715,  1722.  The  gradual  increase  here 
probably  indicates  a  slight  rise  of  temperature. 

Exp.  4.  Partial  Pressures  and  Volumes. 

Apparatus. — All  is  furnished  as  seen  in  the  sketch.  The  absorbing 
solution  consists  of  1  :  1  ammonia  nearly  saturated  with  NH4C1.  The 
copper  may  be  in  the  form  of  turnings  or  in  fine  strips  of  foil  cut  like 
"  excelsior."  The  mixture  thus  prepared  will  absorb  many  liters  of 
oxygen  without  apparent  deterioration.  A  large,  heavy  ring  stand, 
having  a  shelf  attached  to  a  ring,  is  needed  to  support  the  absorption 
tubes. 

Results.— Volume  N2  76.3  cc.;  O2  19.9  cc.;  Ar  0.8  cc.;  H20  2.83  cc. 

Exp.  5.  Charles'  Law. 

Apparatus. — Furnished  as  shown.  The  tube  containing  the  index  is 
kept  in  a  long  thermometer  box,  always  with  the  drying  tube  attached. 
Care  should  be  taken  not  to  shake  it  or  the  index  may  be  separated  into 
several  parts. 

Results. — If  the  bore  of  the  tube  is  uniform  and  no  moisture  gets  in, 
the  results  will  be  good.  Sample  value,  0.00363. 

Exp.  6.  Diffusion. 

Apparatus. — Furnished  as  shown. 

Exp.  7.  Diffusion  and  Molecular  Weights. 

Apparatus. — Furnished  as  shown.  If  any  of  the  gases  are  furnished 
in  cylinders,  reducing  valves  must  be  attached.  If  the  clay  cylinders 
are  frail,  they  may  be  supported  by  tying  a  strong  cord  about  the  open 
end. 

Results. — Time  of  outflow  for  02,  9.84  min. 

Time  of  outflow  for  CO2,  11.23  min. 

«/*=0.87  V32/V5=0.85. 

Time  of  outflow  for  H2,  2.76  min. 

Mol.  wt.  of  H2  =2.48;  calculated  for  H2+ water  vapor,  2.42. 

Exp.  8.  Vapor  Pressure  of  Water. 

Apparatus. — Furnished  as  shown.  Apparatus  must  also  be  pro- 
vided for  heating  the  U-tubes.  Use  a  3-qt.  graniteware  stock  pot 
(13  cm.  deep)  insulated  with  asbestos  paper  and  covered  with  a  thick 
piece  of  asbestos  board  containing  a  slot  for  the  U-tube. 


98  APPENDIX  III 

Wool  is  used  because  wet  cotton  packs  too  closely. 
Results. — A  little  low  always,  because  of  the  imperfect  drying  power 
of  calcium  chloride.     A  sample  value  is  14.81  mm.  instead  of  15  mm. 

Exp.  9.  Heat  of  Evaporation. 

Apparatus. — Furnish  only  the  condenser  with  the  three  connecting 
tubes.  The  transite  board  is  a  part  of  each  student's  equipment. 

The  balance  referred  to  is  a  "  trip  "  with  agate  bearings.  If  prop- 
erly cared  for,  these  balances  will  weigh  closer  than  0.1  gm.  It  is  a  great 
advantage  to  have  on  the  balance  a  12-cm.  evaporating  dish  with  coun- 
terpoise. Students  will  need  some  encouragement  to  use  this,  however; 
they  seem  to  like  to  spill  chemicals  about. 

Ice  is  needed  for  this  experiment  and  for  several  others  in  the  course. 
It  is  a  good  thing  to  have  an  ice-box  in  the  basement  of  the  laboratory 
for  this  and  other  courses,  particularly  the  organic,  and  to  keep  this  filled 
all  the  time. 

Results.— 526  caL,  532  cal. 

Exp.  10.  Molecular  Weight  of  Carbon  Dioxide. 

Apparatus. — A  Kipp  generator  for  C02.  The  tubes  A  and  B  are 
connected  as  seen,  and  attached  permanently  to  a  wooden  base.  The 
student  prepares  the  rest  of  the  apparatus.  The  air  blast  will  be  needed, 
as  in  11. 

Results.— 43.8,  44.1. 

Exp.  11.  Molecular  Weight  of  Ether. 

Apparatus. — Furnish  the  Dumas  bulb  arranged  as  seen,  also  the  bulb 
holder  and  the  thistle  tube.  The  rings  of  the  bulb  holder  should  be 
covered  with  rubber  tubing,  to  prevent  breaking  the  bulb.  Attach 
permanently  to  the  air-blast  line  a  slender  brass  tube  to  dry  out  the  bulb. 

Chemicals. — The  ether  should  be  pure  and  "  dried  over  sodium." 

Results.— 75.8,  75.0. 

Exp.  12.  Composition  of  Silver  Oxide. 
Apparatus. — Student's  outfit. 

Chemicals.— AgNO3;  stick  NaOH,  "  pure  by  alcohol." 
Results.— A  little  low,  Ag  =93.01,  92.8,  93.03. 

Exp.  13.  Composition  of  Silver  Chloride. 

Apparatus. — Outfit. 

Chemicals. — Prepare  suspension  of  asbestos  for  Gooch  crucibles. 
6  NXHC1  will  also  be  needed. 

Results.— Ag  75.36  per  cent,  Cl  24.64  per  cent. 
Ag  75.26  per  cent,  Cl  24.74  per  cent. 


DATA  AND  SUGGESTIONS  99 

Exp.  14.  Multiple  Proportions. 

Apparatus. — Outfit. 

Chemicals. — Suspension  of  asbestos  as  in  13;  HgCl2;  HgCl;  stick 
NaOH,  as  in  12;  N/5  AgN03  (34  gm.  AgNO3  per  liter). 

Results.— Cl  per  1  gm.  Hg;  0.1775  in  HgCl,  0.3528  in  HgCl2. 

Exp.  15.  The  Law  of  Volumes. 

Apparatus. — The  combustion  tubes  are  loaned  to  the  students  for 
the  day  only.  The  students  can  fill  them  easily  by  attaching  a  large- 
bore  funnel-tube  to  one  end  and  then  slowly  shaking  the  oxide  in. 

The  pneumatic  pans  are  made  of  copper;  they  are  12X8  inches 
horizontally,  and  7  in.  deep.  Keep  them  in  the  laboratory  for  common 
use. 

Chemicals. — CuO,  wire  form;  soda-lime;  cone,  ammonia;  long-fiber 
asbestos. 

Results.— N2  167  cc.,  H2  505  cc.,  NH3  334  cc. 

Exp.  16.  Specific  Heat  and  At.  Wt.  of  Tin. 

Apparatus. — Delicate  thermometers,  range  0-50°  in  0.1°.  About 
five  will  do  for  a  class  of  30.  These  may  be  kept  in  the  storeroom,  but 
better  by  the  instructor.  Small  pocket  lenses. 

Chemicals. — Granulated  tin,  30  mesh.  Better  sift  out  the  particles 
as  small  as  60  mesh.  Students  will  heat  the  tin  with  direct  flame  to  dry 
it,  and  will  surely  melt  it  down. 

Results. — 0.0533,  approximate  atomic  weight  120. 

Exp.  17.  Valence  of  Sodium,  Magnesium,  and  Aluminum. 

Apparatus. — Furnish  only  the  capsules  (No.  1). 

Chemicals. — Sodium  under  benzene;  magnesium,  and  aluminum  wire 
No.  12,  cut  in  12-cm.  lengths.  It  is  a  good  thing  to  help  the  students 
when  they  are  cutting  sodium. 

Results.— Cc.  of  H2;  Na  147,  Mg  274,  Al  438. 

Exp.  18.  Oxidizing  and  Reducing  Valence. 

Apparatus. — Outfit. 

Chemicals. — Stock  solution  of  N/10  iodine  (12.7  gm.  I2  ground  up 
with  about  25  gm.  KI,  and  dissolved  in  water  to  make  1  liter); 
Na2S203-5H2O;  starch;  KMnO4;  FeSO4(NH4)2SO4-6H2O;  K2Cr2O7; 
cone,  and  6N  H2SO4;  KI,  10  per  cent  solution. 

Results.—  I2,  25  cc.  =Na2S2O3,  25.1  cc. 

KMnO4,  10cc.=Na2S2O3,  50  cc. 
KMnO4,  9.8  cc.  =FeSO4,  50  cc. 
K2Cr207,  5cc.=Na2S203,  30  cc. 


100  APPENDIX  III 

Exp.  19.  Jones  Reductor. 

Apparatus. — The  Jones  reductor,  fitted  up  as  directed  with  zinc,  etc. 
One  is  enough  for  a  class  of  thirty. 

Chemicals.— FeS04(NH4)2SO4-6H20  and  KMn04,  as  in  18. 

Results. — 50  cc.  iron  solution  required  10  cc.  KMnO4  solution  before 
passing  through  the  reductor  and  10.1  cc.  after. 

Exp.  20.  Supersaturation  of  Sodium  Sulphate. 
Apparatus. — Outfit. 
Chemicals. — Na2S04,  anhydrous;  cotton. 

Exp.  21.  Test  for  Potassium. 
Apparatus. — Outfit. 
Chemicals. — Tartaric  acid;  KNOs. 

Exp.  22.  Normal  Acids. 

Apparatus. — Hydrometers,  range  1-1.2,  with  tall  cylinders  (4^X24 
cm.  inside) ;  50 -cc.  graduated  flasks.  The  flasks  should  be  calibrated 
by  filling  with  an  accurate  pipette  and  remarking. 

Chemicals. — Stock  solutions  of  approximately  6N  acids  (HC1, 
HNOs,  H2SO4).  These  are  tested  out  when  cold  to  see  that  their  gravity 
really  comes  on  the  tables  given. 

Exp.  23.  Normal  Sodium  Hydroxide. 

Apparatus. — Outfit. 

Chemicals. — NaOH,  as  in  12;  methyl  orange  indicator  (0.1  per  cent 
solution  in  water) . 

Results.— NaOH  checks  with  HC1;  25.7  cc.  NaOH  =  25.9  cc.  HNO3; 
24.3  cc.  NaOH  =  24.6  cc.  H2S04. 

Exp.  24.  Freezing-point  Lowering. 

Apparatus. — The  whole  set-up  as  indicated:  five  sets  for  a  class  of 
thirty,  allowing  four  to  work  together.  (Note  thermometer  with 
range  -10  to  +50.) 

Chemicals. — Absolute  ethyl  alcohol;  sal   and  ice  for  refrigeration. 

Results. — Average  of  three  readings,  1.85°. 

Exp.  25.  Molecular  Weight  of  Propyl  Alcohol. 
Apparatus. — As  in  24. 

Chemicals. — Absolute  propyl  alcohol;  salt  and  ice  as  in  24. 
Results.— 61.6,  61.25. 

Exp.  26.  Osmotic  Pressure. 

Apparatus. — Diffusion  thimble,  Schlericher  and  Schull,  No.  579 
(16X100  mm.);  glass  adapter;  capillary  tube ;  tall  cylinder. 


DATA  AND  SUGGESTIONS  lOV 

Chemicals. — Cane  sugar.  Color  the  sugar  solution  with  a  little 
methyl  violet. 

Exp.  27.  Salt  Effect. 

Apparatus. — Outfit. 

Chemicals. — Methyl  orange  indicator,  as  in  23;  N  HC2H3O2 
(57.1  cc.  glacial  acetic  acid,  sp.  gr.  1.055,  per  liter);  pure  neutral  NaCl; 
N  NH4OH  (66  cc.  cone,  ammonia,  sp.  gr.  0.90,  per  liter);  M/5  MgS04 
(49  gm.  MgS04-7H20  per  liter). 

Exp.  28.  lonization  of  Sodium  Chloride. 
Apparatus. — As  in  24. 

Chemicals. — Pure  NaCl,  as  in  27;  salt  and  ice,  as  in  24. 
Results. — 78.3  per  cent — high  value  due  to  hydration. 

Exp.  29.  Chemical  Tests. 

Apparatus. — Outfit. 

Chemicals. — (A)  M.  H3P04  (68  cc.  cone,  acid,  sp.  gr.  1.7,  per  liter). 
Sheet  zinc  in  pieces  1X2  cm.;  brass  wire  gauze,  40  mesh,  pieces  2X4 
cm. ;  N  HC2H3O2,  as  in  27. 

(B)  FeCl3  sol.,  5  per  cent;  NH4SCN  sol., 5  per  cent;  K3Fe(CN)6  sol., 
5  per  cent;    K3Fe(C2O4)3  sol. — Prepare  by  grinding  up  about  5.gm. 
anhydrous  Fe2(S04)3  with  about  10  gm.  K2C2O4,  and  dissolving  in  600 
cc.  water. 

(C)  NaCl  sol.,  5  per  cent;  KC1O3  sol.,  5  per  cent;  AgNO3,  test  sol. 

(D)  H3B03  sol.,  5  per  cent;  methyl  orange,  as  in  23. 

(E)  HgCl2  sol.,  5  per  cent;  Hg(N03)2  sol.,  5  per  cent;  K2Cr2O7  sol., 
5  per  cent. 

Exp.  30.  Hydrolysis  of  Methyl  Acetate. 

Apparatus. — Outfit. 

Chemicals. — Methyl  acetate.  The  commercial  article  usually  con- 
tains methyl  alcohol,  acetone,  and  water.  A  fairly  good  grade  can 
be  made  from  this  by  shaking  with  a  saturated  solution  of  CaC^  and 
redistilling.  The  pure  ester  can  be  obtained  commercially.  M. 
H3PO4  as  in  29;  phenolphthalein,  1  per  cent  sol.  in  alcohol. 

Results. — Cubic  centimeters  of  N  acetic  acid  produced: 

With  HC1  10.1 

"    HNO3  9.62 

"    H2SO4  6.1 

"    H3P04  2.7 


102 


APPENDIX  III 


Exp.  31.  Heat  of  Neutralization. 

Apparatus. — Thermometer,  range  0-50°  in  0.1°. 

Results.— NaOH  and  HC1, 13893  cal;  NaOHandHNOj;  13831  cal. 

Exp.  32.  End  Point  of  Methyl  Orange. 

Apparatus. — Comparator. — This  is  made  of  pine  or  cypress  boards 
5/16  in.  thick,  and  is  of  the  following  dimensions:  length  3  ft.;  ends 
1  ft.  high  by  8  in.  wide;  width  of  bars  3|  in.;  space  between  bars  4  in; 
seventeen  holes  If  in.  in  diameter. 

The  Nessler's  tubes  are  31 X220  mm. ;  100-cc.  mark  45  mm.  from  the 
top ;  bottoms  cut  flat. 

Chemicals. — Methyl  orange,  as  in  23. 

Results.— 2  X10-4,  always. 

Exp.  33.  End  Point  of  Phenolphthalein. 
Apparatus. — Same  as  in  32. 
Chemicals. — Phenolphthalein  indicator,  as  in  30. 
Results. — Rather  poor. 

Exp.  34.  Choice  of  Indicator. 
Apparatus. — Outfit. 
Chemicals. — N  acetic  acid,  as  in  27;  N  NH40H,  as  in  27. 

Exp.  35.  Titration  of  Polybasic  Acids. 

Apparatus. — Outfit. 

Chemicals.— M  H3P04,  as  in  29;  NaHC03;  Na2C03  anhydrous; 
indicators  as  before. 


Exp.  36.  Speed  of  Reaction. 

Apparatus. — Outfit. 

Chemicals. — Methyl  acetate,  as  in  30;  phenolphthalein  indicator,  as 
in  30. 

Results.— 


Relative  speed  calc. 

Relative  speed  obs. 

K. 

1 

1 

10.5 

1.93 

1.78 

9.8 

2 

2.03 

10.7 

3.87 

3.53 

9.7 

DATA  AND  SUGGESTIONS  103 

Exp.  37.  Equilibrium  Constant. 

Apparatus. — Outfit. 

Chemicals. — Methyl  acetate,  as  in  30;  glacial  acetic  acid;  pure 
anhydrous  methyl  alcohol. 

Results. — The  two  values  for  K  were,  in  different  cases: 

7.3  and  8.4 
6.9  and  7.3 
8.7  and  8.4,  the  value  depending  on  the  temperature. 

Exp.  38.  Ionic  Equilibrium. 
Apparatus. — Outfit. 
Chemicals.— CuBr2;  NaBr;  Cd(N03)2. 

Exp.  39.  Common  Ion  Effect. 

Apparatus. — Outfit. 

Chemicals.— M  acetic  acid,  as  in  27;  NaC2H302-3H2O;  N/10 
NH4OH. — Students  prepare  from  M  (27);  NH4C1;  methyl  orange  and 
phenolphthalein. 

Results. — Correspond  exactly  with  calculated  values. 

Exp.  40.  Neutralization  Effect. 

Apparatus. — Outfit. 

Chemicals. — NaC2Hs02,  as  in  39;  methyl  orange;  congo  red 
(1  per  cent  sol.  in  30  per  cent  alcohol);  NEUCl,  as  in  39;  thymolphtha- 
lein  indicator  (1  per  cent  sol.  in  alcohol). 

Results. — As  calculated. 

Exp.  41.  Hydrolysis. 

(A)  Apparatus. — Outfit. 

Chemicals. — Five  per  cent  solutions  of  NaC2Hs02,  K2COs,  Aids, 
NaCl,  NH4C1;  red  and  blue  litmus  solutions  (extract  20  gm.  of  the 
powdered  cubes  twice  with  200  cc.  hot  water.  Let  stand  in  tall  beaker 
to  settle  and  decant  clear  liquid  through  a  Buchner.  Make  up  to  600  cc. 
divide  into  two  portions,  and  redden  one  portion  with  a  drop  or  two 
of  HC1). 

(B)  Apparatus. — Comparator  and  Nessler's  tubes. 
Chemicals. — Aniline  sulphate;  methyl  orange. 
Results.— Calculated  value,  0.0151;  observed,  0.0115. 

Exp.  42.  Dehydration  of  Copper  Sulphate  Pentahydrate. 
Apparatus. — Outfit. 
Chemicals.— CuSO4  •  5H20. 
Results.— Loss  at  200°,  3.97  mol.  H20;  at  250°,  4.96  mol.  H2O. 


104  APPENDIX  III 

Exp.  43.  Partition  of  Bromine. 

Apparatus. — 100-cc.  glass-stoppered  bottles;  about  1  dozen  pairs 
will  be  needed  for  a  class  of  thirty.  These  are  best  kept  by  the  in- 
structor. 

Chemicals. — Bromine  water,  cone.;  10  per  cent  sol.  of  KI,  as  in  18; 
Na2S203,  as  in  18;  CC14. 

Results. — Values  for  K  in  the  two  cases,  25.1,  26. 

Exp.  44.  Partition  of  Succinic  Acid. 

Apparatus. — Same  as  in  43. 

Chemicals.— M/ 10  succinic  acid  (about  12  gm.  (CH2COOH)2  per 
liter) ;  dry  ether. 

Results. — K  in  the  two  cases,  6.8  and  6.9. 

Exp.  45.  Phase  Rule. 

Apparatus. — Outfit,  reading  lens. 

Chemicals.— Na2SO4  •  10H20. 

Results. — Very  exact;  the  non-variant  system  continues  for  about 
two  hours. 

Exp.  46.  Precipitation  of  Silver  Acetate. 

Apparatus. — Outfit. 

Chemicals. — N/5  sodium  acetate  (27.2  gm  NaC2H3O2-3H2O  per 
liter);  N/10  AgN03  (17  gm.  AgN03  per  liter);  N/5  AgN03,  as  in  14 

Exp.  47.  Precipitation  of  Sulphides. 

Apparatus. — Outfit;  an  H2S  generator  which  the  student  may 
make  for  himself. 

Chemicals.— M/10  ZnS04  (29  gm.  ZnS04-7H2O  per  liter);  FeS, 
granulated;  NH4OH;  sodium  acetate,  as  in  39;  M/10  CuS04  (25  gm. 
CuS04-5H2O  per  liter);  sheet  zinc,  as  in  29;  sodium  acetate  as  in  39; 
K4Fe(CN)6,  5  per  cent;  FeS04(NH4)2S04-6H20,  as  in  18. 

Exp.  48.  Precipitation  of  Magnesium. 

Apparatus. — Outfit. 

Chemicals.— M  NH4OH,  as  in  27;  M/5  MgS04,  as  in  27;  NH4C1; 
Na2HPO4,  5  per  cent  sol.' 

Exp.  49.  Silver-ammonium  Complex. 

Apparatus. — Outfit . 

Chemicals.— N/5  AgNO3,  as  in  14;  M/5  NH4OH— student  pre- 
pares from  M  NH4OH  (27);  N/10  NaCl  (approximately  5.8  grn.  per 
liter);  N.  NH4OH  as  in  27;  N/10  KBr  (approximately  12  gm.  per 
liter). 

Results. — As  calculated. 


DATA  AND  SUGGESTIONS  105 

Exp.  50.  Ferric-oxalate  Complex. 

Apparatus. — Outfit. 

Chemicals.— Anhydrous  Fe2 (S04)3;  (NH4)2  C2O4-2H2O;  NH4SCN 
sol.,  as  in  29;  CaCl2  sol.,  5  per  cent;  FeCls,  granulated. 

Exp.  51.  Amphoterism. 

Apparatus. — Small  nickel  crucibles. 

Chemicals.— 5  per  cent  A12(S04)3;  dilute  NaOH  and  NH4OH  pre- 
pared by  student;  5  per  cent  Cr2(SO4)a;  5  per  cent  FeCla  sol.  as  in  29; 
Na202;  Fe20s;  5  per  cent  BaCb  sol. 

Exp.  52.  Amphoteric  Bromine. 
Apparatus. — Outfit. 
Chemicals. — Sol.  of  KI,  as  in  18;  bromine  water,  as  in  43;  KBrOs;  I2. 

Exp.  53.  Faraday's  Law. 

Apparatus. — Coulometer  with  two  copper  electrodes;  ammeter  of 
low  range;  rheostat,  26  ohms;  connecting  wires. 

Chemicals. — Special  copper  sulphate  solution,  see  Exp.;  alcohol, 
96  per  cent. 

Exp.  54.  Electrode  Reactions. 

Apparatus. — U-tube  with  two  platinum  electrodes  and  one  copper 
electrode ;  wires. 

Chemicals. — Na2SO4,  5  per  cent;  FeS04,  5  per  cent;  FeCls; 
K3Fe(CN)6,  and  NH4SCN  as  in  29. 

Exp.  55.  Migration  of  Hydrogen  and  Hydroxyl  Ions. 
Apparatus. — Duplicate  of  54. 

Chemicals. — Agar-agar  (powdered);  sat.  sol.  of  KNOs. 
Results. — Low,  but  relatively  correct:  H+  7.03  cm.     OH~  3.75  cm. 

Exp.  56.  Migration  of  Copper  ions. 

Apparatus. — Special  U-tube  with  electrodes  and  wires. 
Chemicals. — Sol.  of  CuBr2  (1  gm.  CuBr2  :  2|  gm.  water). 

Exp.  57.  Daniell  Cell. 

Apparatus. — Daniell  cell  with  sheet-copper  electrode;  wires;  volt- 
meter and  ammeter  of  low  range. 

Chemicals.— N  ZnS04  (144  gm.  ZnS04-7H2O  per  liter);  saturated 
CuS04;  cone.  ZnCl2;  N  CuS04  (125  gm.  CuSO4-5H20  per  liter);  sol. 
of  KCN,  5  per  cent. 

Students  are  very  likely  to  get  copper  solution  inside  the  porous  cup. 
This  deposits  copper  on  the  zinc  rods  and  then  brings  about  local  action 


106  APPENDIX  III 

and  change  of  voltage.     The  rods  should  be  frequently  cleaned  (scraped 
or  sandpapered). 

Results. — Close  to  calculated  values. 

Exp.  58.  Concentration  Cell. 

Apparatus. — Same  as  in  57,  but  with  sheet-zinc  electrode. 

Chemicals. — 0.01  N  ZnSO4,  prepared  by  student  from  N;  cone. 
ZnCl2,  as  in  57. 

Exp.  59.  Decomposition  Voltage. 

Apparatus. — Coulometer  cell;  platinum  electrodes;  voltmeter  and 
ammeter  of  low  range;  2200-ohm  rheostat;  mercury  to  cover  bottom 
of  cell;  connecting  spiral  of  rubber-insulated  copper  wire. 

Results. — Almost  theoretical.  Students  are  inclined  to  rush  this 
experiment  through  and  then  to  draw  very  poor  curves. 

Exp.  60.  Ionic  Displacement. 

Apparatus. — Outfit. 

Chemicals. — SnCU  sol.,  5  per  cent;  sheet  zinc  in  strips  1X2  cm.; 
iron  wires  No.  18,  30  cm.  long;  SnCk  sol.,  5  per  cent;  SbCla  sol.,  5  per 
cent;  EbS  solution  (take  a  2-liter  brown  bottle,  fill  three-fourths  full  of 
water,  add  10  cc.  cone.  NH4HS  sol.,  and  acidify  with  HC1. — keep 
corked);  strips  of  zinc,  iron,  and  lead,  1X6  cm.;  mercury.  (Do  not  let 
students  put  zinc  in  the  clean  mercury  used  in  59.) 

Results.— Perfect, 


APPENDIX  IV 
Chemicals 

The  following  list  includes  all  the  chemicals  mentioned  in  Appendix 
III,  arranged  alphabetically. 

(A)  Solutions  and  Liquids : 

Acetic  acid  N  (57.1  cc.  glacial  acetic  acid,  sp.  gr.  1.055  per  liter). 

Acetic  acid,  glacial. 

Alcohol,  absolute  (not  on  the  side  shelf) 

Alcohol,  96  per  cent. 

Aluminum  chloride,  5  per  cent. 

Aluminum  sulphate,  5  per  cent. 

Ammonium  chloride,  5  per  cent. 

Ammonium  hydroxide,  cone.,  sp.  gr.  0.90. 

Ammonium  hydroxide,  N  (66  cc.  cone.  NH4OH,  sp.  gr.  0.90,  per 

liter). 

Ammonium  thiocyanate,  5  per  cent. 
Antimony  trichloride,  5  per  cent. 
Asbestos  suspension  (as  in  13). 
Barium  chloride,  5  per  cent. 
Boric  acid,  5  per  cent. 
Bromine  water,  cone. 
Calcium  chloride,  5  per  cent. 
Carbon  tetrachloride. 
Chromium  sulphate,  5  per  cent. 
Congo  red  indicator,  1  per  cent  sol.  in  alcohol. 
Cupric  bromide  (1  :  2|). 
Cupric  sulphate,  saturated. 

Cupric  sulphate,  M/10  (25  gm.  CuSO4-5H2O  per  liter). 
Cupric  sulphate  N  (125  gm.  CuSO4-5H2O  per  liter). 
Ether,  U.  S.  P. 

Ether,  dried  over  sodium  (not  on  the  side  shelf). 
Ferric  chloride,  5  per  cent. 
Ferrous  sulphate,  5  per  cent. 
Hydrochloric  acid,  cone.  sp.  gr.  1.19. 

Hydrochloric  acid,  6N  (dilute  cone,  with  equal  volume  of  water). 

107 


108  APPENDIX  IV 

Hydrogen  sulphide,  saturated  sol.  (as  in  60) . 

Iodine  N/10  sol.  in  KI  (12.7  gm.  I2,  25  gm.  KI  per  liter). 

Litmus,  red  (as  in  41). 

Litmus,  blue  (as  in  41). 

Magnesium  sulphate,  M/5  (49  gm.  MgSO4-7H2O  per  liter). 

Mercuric  chloride,  5  per  cent. 

Mercuric  nitrate,  5  per  cent. 

Mercury. 

Methyl  acetate,  pure,  anhydrous. 

Methyl  alcohol,  pure,  anhydrous. 

Methyl  orange,  0.1  per  cent  sol.  in  water. 

Nitric  acid,  cone.,  sp.  gr.  1.42. 

Nitric  acid,  6N  (dilute  400  cc.  cone.  HN03,  sp.  gr.  1.42  to  1  liter), 

Phenolphthalein,'!  per  cent  sol.  in  alcohol. 

Phosphoric  acid,  M  (68  cc.  cone,  acid,  sp.  gr.  1.7,  per  liter). 

Potassium  bromide,  M/10  (12  gm.  KBr  per  liter). 

Potassium  carbonate,  5  per  cent. 

Potassium  chlorate,  5  per  cent. 

Potassium  cyanide,  5  per  cent. 

Potassium  dichromate,  5  per  cent. 

Potassium  ferricyanide,  1  per  cent  (fresh). 

Potassium  iodide,  10  per  cent. 

Potassium  nitrate,  saturated  sol. 

Potassium  ferric-oxalate  (as  in  29,  must  be  fresh). 

Potassium  ferrocyanide,  5  per  cent. 

Propyl  alcohol,  pure,  anhydrous  (not  on  the  side  shelf). 

Silver  nitrate,  test  sol. 

Silver  nitrate,  N/10  (17  gm.  AgNO3  per  liter). 

Silver  nitrate,  N/5  (34  gm.  AgN03  per  liter). 

Sodium  acetate,  5  per  cent. 

Sodium  acetate  N/5  (27.2  gm.  NaC2H3O2 •  3H20  per  liter). 

Sodium  chloride  N/10  (5.8  gm.  NaCl  per  liter). 

Sodium  chloride,  5  per  cent. 

Sodium  hydroxide  N/2  (standardized). 

Sodium  phosphate,  di.,  5  per  cent. 

Sodium  sulphate,  5  per  cent. 

Stannic  chloride,  5  per  cent. 

Stannous  chloride,  5  per  cent. 

Succinic  acid,  M/10  (12  gm.  (CH2COOH)2  per  liter). 

Sulphuric  acid,  cone.,  sp.  gr.  1.84. 

Sulphuric  acid,  6  N  (166  cc.  cone,  acid,  sp.  gr.  1.84,  per  liter). 

Thymolphthalein,  1  per  cent  sol.  in  alcohol. 


CHEMICALS  109 

Zinc  chloride,  cone. 

Zinc  sulphate  M/10  (29  gm.  ZnSO4-7H2O  per  liter). 

Zinc  sulphate,  N  (144  gm.  ZnS04-7H20  per  liter). 

(B)  Solids: 

Agar-agar,  powdered. 

Aluminum  wire,  No.  12  (in  15-cm.  lengths). 

Ammonium  chloride. 

Ammonium  oxalate. 

Aniline  sulphate. 

Asbestos,  long  fiber. 

Asbestos,  for  Gooch  crucibles. 

Brass  wire  gauze,  40-mesh  (in  pieces  2X4  cm.). 

Cadmium  nitrate. 

Calcium  chloride,  anhydrous. 

Cane  sugar. 

Copper  turnings. 

Cupric  bromide. 

Cupric  oxide,  wire  form. 

Cupric  sulphate. 

Ferric  chloride. 

Ferric  oxide. 

Ferric  sulphate. 

Ferrous  ammonium  sulphate. 

Ferrous  sulphide. 

Gamboge. 

Glass  beads,  2  mm. 

Iodine. 

Iron,  strips  (1X6  cm.). 

Iron  wire,  No.  18  (in  30-cm.  lengths). 

Lead  strips  (1X6  cm.). 

Magnesium  wire,  No.  12  (in  15-cm.  lengths). 

Mercuric  chloride. 

Mercurous  chloride. 

Potassium  bromate. 

Potassium  dichr ornate. 

Potassium  nitrate. 

Potassium  permanganate. 

Salt,  for  refrigerating. 

Silver  nitrate. 

Soda  -lime,  granulated. 

Sodium. 


110  APPENDIX  IV 

Sodium  acetate. 

Sodium  bicarbonate. 

Sodium  bromide. 

Sodium  carbonate  (anhydrous) . 

Sodium  chloride,  pure,  neutral. 

Sodium  hydroxide,  pure  by  alcohol. 

Sodium  peroxide. 

Sodium  sulphate,  anhydrous. 

Sodium  sulphate,  decahydrate. 

Sodium  thiosulphate. 

Starch,  potato. 

Tartaric  acid. 

Tin,  granulated,  30-mesh. 

Zinc,  strips  (1X6  cm.). 


APPENDIX  V 
Apparatus 

(A)  Student's  Outfit. — Each  student  is  furnished  with  the   following 
articles  which  he  keeps  during  the  year: 

1  Air  bath,  graniteware. 

6  Beakers,  2  each  of  500,  300  and  100  cc. 

5  Bottles,  glass-stoppered,  three  600  cc.,  two  1000  cc. 

2  Bunsen  burners. 

2  Burettes,  1  Mohr,  1  stop-cock. 
1  Biichner  funnel,  9  cm. 

1  Camel's  hair  brush. 

2  Casseroles,  300  cc.  and  125  cc. 

2  Crucibles,  No.  0  and  No.  1,  with  covers. 
1  Gooch  crucible  with  disk. 

1  Desiccator,  6-inch. 

2  Evaporating  dishes,  7  and  9  cm. 
1  Filter  stand. 

1  Filter  flask,  1  liter. 

2  Flasks,  graduated,  500  cc.  and  1000  cc. 
2  Flasks,  Florence,  500  cc.  and  1500  cc. 
2  Flasks,  Erlenmeyer,  250  and  500  cc. 

4  Funnels,  6  cm.,  long  stem. 
1  Graduated  cylinder,  50  cc. 
1  Key  for  locker. 

1  Mortar,  10-cm.  porcelain,  with  pestle. 

5  Pipettes,  1,  5,  10,  50,  and  100  cc. 

2  Ring  stands,  2  clamps,  and  1  four-inch  iron  ring. 
1  Test-paper  bottle. 

6  Test-tubes,  medium  size. 

1  Test-tube,  25X180  mm.,  heavy. 
1  Tile,  glazed  porcelain,  6-inch. 
1  Tongs,  brass. 

1  Transite  board,  12X12  inch. 
1  Tripod,  iron. 

Ill 


112  APPENDIX  V 

1  Water  bath,  graniteware  with  1   galvanized  iron  ring  and  4 

porcelain  rings.* 

2  Watch  glasses,  7  and  10  cm. 
1  Weighing  bottle. 

1  Waste  jar. 

(B)  Non-returnable  Articles : 

The  following  articles  are  ordered  from  the  storeroom  as  needed. 
They  cannot  be  returned  except  in  special  cases  which  must  be  decided 
-by  the  instructor: 

Clay  triangles  Rubber  stoppers 

Files  Rubber  tubing 

Filter  paper  Sponges 

Glass  rods  Towels 

Glass  tubing  Test-tube  brushes 

Matches  Tube  brushes 

Wire  gauze 

(C)  Special  Apparatus : 

There  are  many  other  pieces  of  apparatus  not  included  in  the  above 
lists.  Most  of  these  are  a  part  of  the  common  stock  of  any  chemical 
laboratory.  Any  very  special  apparatus  is  either  indicated  in  the 
sketches  or  is  mentioned  in  the  accompanying  descriptions. 

*  A  3-quart  graniteware  stock  pot  with  perpendicular  sides  used  also  as  indicated 
in  Exp.  11,  etc.  The  iron  ring  serves  as  an  adapter.  It  is  turned  over  the  edge 
of  the  pot  and  fits  the  largest  porcelain  ring. 


APPENDIX  VI 
LOGARITHMS. 


Natural  I 
Number*,  j 

0 

1 

2 

3 

4 

5 

6 

7 

8 

PRO  FOR 

9 

noNAL,  PARTS. 

133- 

56789 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374  4    8  12 

21  25  29  33  37 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755  4  8  ll 

19  23  26  30  34 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106       7  10 

17  21  24  28  31 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430       6  10 

16  19  23  26  29 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732       6    9 

15  18  21  24  27 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014       6    81 

14  17  20  22  25 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279       5    81 

13  16  18  21  24 

17 

23Q4 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529       5    7  l 

12  15  17  20  22 

18 

2553 

2577 

2601 

2625'  2648 

2672 

2695 

2718 

2742 

2765       5    7 

12  14  16  19!21 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989       4    7 

11  13  16  18  20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201       4    6 

11  13  15  17  19 

21 

3222 

3243 

3263 

32843304 

3324 

3345 

3365 

3385 

3404       4    6 

1012141618 

22 

3424 

3444 

3464 

34833502 

3522 

3541 

3560 

35793598       4    6 

10  12  14  15  17 

23 

3617 

3636 

3655 

3674  3692 

3711 

3729 

3747 

37663784       4    6 

911  131517 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962  245 

911  121416 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133  235 

9  10  12  14  15 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298  235 

8  10  11  13  15 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456  235 

8    9  11  13  14 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

45944609  235 

8    9  11  12  14 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757  i   3    4 

7    9101213 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900  i   3    4 

7    9  10  11  13 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038  i   3    4 

7    8101112 

32 

5051 

5065 

5079 

50925105 

5119 

5132 

5145 

5159 

5172  i   3    4 

7    8    91112 

33 

5185 

5198 

5211 

5224'  5237 

5250 

5263 

5276 

5289 

5302  i   3    4 

6    8    91012 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428  i   3    4 

6    8    91011 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551  124, 

6   7    91011 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670  124, 

6         81011 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

6786  123, 

6         8    910 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899  123, 

6         8    9  10 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010  123- 

5         8    910 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117  l   2    3  < 

5    6    8    910 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222  l   2    3  < 

56789 

42 

6232 

6243 

6253 

6263  6274 

6284 

6294 

6304 

6314 

6325  1   2    3  < 

56789 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425  l  2    3  < 

6789 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522  123' 

6789 

45 

6532 

6542 

6551 

6561 

6571 

65SO 

6590 

6599 

6609 

6618  123' 

6789 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712  123' 

C   7   7   8 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803  123' 

5678 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893  l   2    3  ^ 

5673 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981  123^ 

45678 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067  1232 

45673 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152   1    2     3    c 

45678 

52 

^leo 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235  1223 

4567V 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316  1223 

4566? 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396  1223 

45667 

114 


APPENDIX  VI— Continued 
LOGARITHMS. 


[  Natural  I 
Numbera.  1 

0 

1 

2 

3 

4 

5 

C 

7 

8 

9 

PROPORTIONAL  PARTS. 

1 

'^ 

o 

4 

r, 

G 

7 

8 

G 

55 

740^ 

741$ 

I74K 

7427 

7435 

744S 

7451 

745S 

7466 

7474 

1 

2 

„ 

„ 

4 

5 

P 

G 

7 

56 

748r 

7490  7497 

750£ 

7513 

752C 

7528 

7536 

7543|7551 

] 

o 

ri 

2 

5 

4 

5 

5 

6 

7 

57 

755£ 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619  7627 

1 

r\ 

2 

3 

4 

5 

5 

6 

7 

58 

7834 

7642 

764S 

7857 

7664 

7672 

7679 

7686 

7694  7701 

1 

i 

2 

3 

•1 

4 

5 

8 

7 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

1 

i 

2 

3 

4 

4 

5 

6 

7 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

1 

i 

2 

3 

4 

4 

5 

6 

5 

61 

7853 

7860 

7888 

7875 

7882 

7889 

789G 

7903 

7910 

7917 

1 

i 

2 

3 

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5 

6 

6 

62 

7924 

7931 

7938 

7945 

7952 

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7966 

7973 

7980 

7987 

1 

i 

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3 

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6 

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63 

7993 

8000 

8007  8014 

8021 

8028 

8035 

8041 

8048 

8055 

1 

i 

2 

3 

3 

4 

5 

5 

6 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

1 

i 

2 

3 

3 

4 

5 

5 

6 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

1 

i 

2 

3 

3 

4 

5 

5 

6 

66 

8195 

8202 

8209 

8215 

8222 

8228  8235 

8241 

8248 

8254 

1 

i 

2 

3 

3 

4 

5 

5 

6 

67 

8261 

8267 

8274 

8280 

8287 

8293  8299 

8306  8312 

8319 

1 

i 

2 

3 

3 

4 

/> 

5 

6 

68 

325 

8331 

8338 

8344 

8351 

8357  8363 

8370 

8376 

8382 

1 

i 

2 

3 

3 

4 

4 

5 

6 

69 

388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

1 

i 

2 

2 

3 

4 

4 

r> 

8 

70 

451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

1 

i 

2 

o 

3 

4 

4 

r> 

6 

71 
72 

513  8519 
573  8579 

8525 

8585 

8531 
8591 

8537 

8597 

8543 
8603 

8549 
8609 

8555 
8615 

8561 
8621 

8567 
8627 

1 

i 
i 

2 
2 

2 
2 

3 
3 

4 

4 

4 

4 

r> 
5 

5 
5 

73 

633  8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

2 

2 

3 

4 

4 

5 

5 

74 

692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

2 

2 

3 

4 

4 

5 

5 

75 

751 

8756 

8762 

8768 

8774 

S779 

8785 

8791 

8797 

8802 

2 

2 

3 

3 

4 

5 

5 

76 

808 

8814 

8820 

8825 

8831 

S837 

8842 

8848 

8854 

8859 

2 

2 

3 

3 

4 

5 

5 

77 

865 

8871 

8876 

8882 

8887 

893 

8899 

8904 

8910 

8915 

2 

o 

3 

3 

4 

4 

5 

78 

921 

8927 

8932 

8938 

8943 

949 

8954 

8960 

8965 

8971 

2 

2 

3 

3 

4 

4 

5 

79 

976 

8982 

8987 

8993 

8998 

004 

9009 

9015 

9020 

9025 

1 

2 

2 

3 

3 

4 

4 

5 

80 

031 

9036 

9042 

9047 

9053 

058 

9063 

9069 

9074 

9079 

1 

i 

2 

2 

3 

3 

4 

4 

5 

81 

085 

9090 

9096 

9101 

9106 

112 

9117 

9122 

9128 

9133 

1 

i 

o 

2 

3 

3 

4 

4 

5 

82 

00 

138 

9143 

Q1  Qf\ 

9149 

Q901 

9154 

Q90R 

9159 

QO1O 

165 
917 

9170 

QOOO 

9175 
0007 

9180 
QOQO 

9186 

QOQQ 

1 

i 

2 

2 

3 

3 

4 

4 

5 

OO 

84 

243 

yiyo 
9248 

y^ui 
9253 

*y£\J\j 

9258 

\j£iL£ 

9263 

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269 

\J  &£t& 

9274 

"  ~.  —  i 

9279 

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9284 

y^oo 

9289 

1 

l 

2 

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85 

294 

9299 

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9315 

320 

9325 

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9340 

1 

l 

2 

2 

3 

3 

4 

4 

5 

86 

345 

9350 

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9360 

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370 

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9380 

9385  9390 

1 

l 

2 

2 

3 

3 

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4 

5 

87 

395 

9400 

9405  9410 

9415 

420 

9425 

9430 

9435  9440 

0 

l 

1 

2 

2 

3 

3 

4 

4 

88 

445 

9450 

9455  9460 

9465 

469 

9474 

9479 

9484  9489 

0 

i 

1 

2 

2 

3 

3 

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4 

89 

494 

9499 

9504 

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9513 

518 

9523 

9528 

95339538 

0 

l 

1 

2 

2 

3 

3 

4 

4 

90 

542 

9547 

9552 

9557 

9562 

566 

9571 

9576 

958119586 

0 

l 

1 

2 

2 

3 

3 

4 

4 

91 

590 

9595 

9600 

9605 

9609 

614 

9619 

9624 

96?8 

9633 

0 

i 

1 

2 

2 

3 

3 

4 

4 

92 

638 

9643 

9647 

9652 

9657 

661 

9666 

9671 

9675 

9680 

0 

i 

1 

2 

2 

3 

3 

4 

4 

93 

685 

9689 

9694 

9699 

9703 

708 

9713 

9717 

9722 

9727 

0 

l 

1 

2 

2 

3 

3 

4 

4 

94 

731 

9736 

9741 

9745 

9750 

754 

9759 

9763 

9768 

9773 

0 

l 

1 

2 

2 

3 

3 

4 

4 

95 

777 

9782 

9786 

9791 

9795 

800 

9805 

9809 

9814 

9818 

0 

l 

1 

2 

3 

3 

4 

4 

96 

823 

9827 

9832 

9836 

9841 

845 

9850  9854  9859 

9863 

0 

i 

1 

2 

3 

3 

4 

4 

97 

868 

98729877 

9881 

9886 

890 

9894  9899i  9903 

9908 

0 

i 

1 

2 

3 

3 

4 

4 

98 

912 

9917  9921 

9926;9930 

934  9939  9943  9948 

9952 

0 

l 

1 

2 

3 

3 

4 

4 

99 

956 

9961  9965  9969 

9974 

9978  9983 

9987 

9991 

0996 

0 

i 

1 

2 

3 

3 

3 

4 

1 

115 


539G    T 


C  tf 


UNIVERSITY  OF  CAUFORNIA  LIBRARY 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


REC'D  LD 

JUL  6    196, 


EC'D  LD 

\  '64-8  PM 


APPENDIX  vn  2  1 564 

INTERNATIONAL  ATOMIC  WEIGHTS,  1921 


Symbol 

Atomic 
weight 

Symbol 

Atomic 
weight 

Aluminium.  .  . 

Al 
Sb 
A 
As 
Ba 
Bi 
B 
Br 
Cd 
Ca 
C 
Ce 
Cs 
Cl 
Cr 
Co 
Cb 
Cu 

Dy 

Er 
Eu 
F 
Gd 
Ga 
Ge 
Gl 
Au 
He 
Ho 
H 
In 
I 
Ir 
Fe 
Kr 
La 
Pb 
Li 
Lu 
Mg 
Mn 
Hg 
Mo 

27.0 

120.2 
39.9 
74.96 
137.37 
209.0 
10.9 
79.92 
112.40 
40.07 
12.005 
140.25 
132.81 
35.46 
52.0 
58.97 
93.1 
63.57 
162.5 
167.7 
152.0 
19.0 
157.3 
70.1 
72.5 
9.1 
197.2 
4.00 
163.5 
1.008 
114.8 
126.92 
193.1 
55.84 
82.92 
139.0 
207.20 
6.94 
175.0 
24.32 
54.93 
200.6 
96.0 

Neodymium  
Neon  

Nd 

Ne 
Ni 

Nt 
N 
Os 
O 
Pd 
P 
Pt 
K 
Pr 
Ra 
Rh 
Rb 
Ru 
Sa 
Sc 
Se 
Si 
Ag 
Na 
Sr 
S 
Ta 
Te 
Tb 
Tl 
Th 
Tm 
Sn 
Ti 
W 
U 
V 
Xe 

Yb 
Yt 
Zn 
Zr 

144.3 
20.2 
58.68 

222.4 
14.008 
190.9 
16.00 
106.7 
31.04 
195.2 
39.10 
140.9 
226.0 
102.9 
85.45 
101.7 
150.4 
45.1 
79.2 
28.3 
107.88 
23.00 
87.63 
32.06 
181.5 
127.5 
159.2 
204.0 
232.15 
169.5 
118.7 
48.1 
184.0 
238.2 
51.0 
130.2 

173.5 
89.33 
65.37 
90.6 

Antimony  
Argon  

Nickel  

Arsenic 

Niton  (radium  em- 
anation). 

Barium 

Bismuth  

Nitrogen  

Boron  
BrominlL 

Osmium  . 

Oxygen 

Cadmium 

Palladium  
Phosphorus  
Platinum 

Calcium  
Carbon 

Cerium  

Potassium    

Cesium 

Praseodymium  .... 
Radium  
Rhodium.  . 

Chlorine  

Chromium  
Cobalt  

Rubidium.  . 

Columbium  

Ruthenium  
Samarium  
Scandium  
Selenium 

Copper.  )v  ;  
Dysprosium.       "*  '" 

Erbium 

Europium  

Silicon  

Fluorine 

Silver.  .  .  . 

Gadolinium  
Gallium  

Sodium 

Strontium  

Germanium  
Glucinum  l  
Gold  

Sulfur  
Tantalum  
Tellurium  

Helium   .... 

Terbium.  .  .  . 

Holmium  
Hydrogen  

Thallium  
Thorium  

Indium  

Thulium     

Iodine       

Tin 

Indium 

Titanium 

Iron  

Tungsten  

Krypton  
Lanthanum 

Uranium  

Vanadium 

Lead  

Xenon  
Ytterbium  
(Neoytterbium)  . 
Yttrium  
Zinc  
Zirconium  

Lithium  

Lutecium 

Magnesium  

Manganese  

Mercury  
Molybdenum  

i  Also  called  Beryllium,  Be. 


[i  ill 


iihii    5   iii 


hi 


